Crack Paths 2009

W1 I 2 UE + W f

(8)

where W1 is the work / energy performed by the applied loads, UE is the energy

dissipated by the growth of tensile cracks, and W}. is the frictional dissipation energy

due to the sliding of the initial crack.

Under fixed loading conditions, the work done by the applied loads 0'1 ,02 on the

additional displacements is

W1 : 4hh(o'1A$1 +o'2Ae2)

(9)

where 4bh is the volume of a unit thickness cell, and A5,,As2 are the strain

increments in e, and 52 due to each crack, respectively.

Based on the linearity of the problem under consideration, the strains increments

(Ag1 and A52) due to the growth of the sliding crack are assumed to be linearly

dependent on the applied compressive stress 61 and 0-2 as

A5; S21 S22 62

where Sij(i, j : 1,2) are constants and S12 I S21.

The workdone by the applied loads is then

W1 : 4bh(Sno-12 + 26162812 + $22622)

(11)

There are two crack tips in the model. The strain energy dissipated at each crack tip

can be expressed by a stress intensity factor KI

l

U6 : ZUC : Z I W K I Z C ”(12)

0

4E

Substituting Eq. (1) into Eq. (12) and then integrating it, the strain energy can be given

as

I(k+1)(1+v){4c2(r*)2sin201n[l+l*]

1

+—U227Z'l2 2

U

e

l

2 E

7r

*

(13)

—2cr*o'2sin6 ln L + 1 + L — L 1 + L

l*

l*

La

La

Under the shear stress ignoring the crack spacing the sliding displacement 5 can be

written as [7]

_(k+l)(l+v) 2cr*sin6_ @ w i n g i n g a,\/E+a,\/;]./2/t(i+i,,)

(14)

Then frictional dissipation energy due to sliding of the initial crack is then

698