Page 57 - T. Anderson-Fracture Mechanics - Fundamentals and Applns.-CRC (2005)

1656_C02.fm Page 37 Thursday, April 14, 2005 6:28 PM

Linear Elastic Fracture Mechanics 37

Equation (2.31) is demonstrated graphically in Figure 2.7(b) and Figure 2.8(b). In load control,
a crack extension da results in a net increase in strain energy because of the contribution of the
external force P:

( dU) = Pd − ∆ Pd∆ = Pd∆
P
2 2
When displacement is fixed, dF = 0 and the strain energy decreases:

( dU) = ∆ dP
∆
2
where dP is negative. As can be seen in Figure 2.7(b) and Figure 2.8(b), the absolute values of
these energies differ by the amount dPd∆/2 , which is negligible. Thus
(dU ) =− (dU ) ∆
P

for an increment of crack growth at a given P and ∆.

EXAMPLE 2.2

Determine the energy release rate for a double cantilever beam (DCB) specimen (Figure 2.9)

Solution: From beam theory

∆ = Pa 3 I = Bh 3
2 3EI where 12

FIGURE 2.9 Double cantilever beam (DCB) specimen.