Crack Paths 2012

Further, outside of the reverse cyclic plastic zone since the constitutive behavior of

the material is supposed to be elastic, it is expected in first approximation that the basic

equations of thermo-elasticity will govern. The equilibrium equation is:

6 0 r

r

+or—o6=0

(7)

0r

For which 0, is the radial normalstress and 09 is the circumferential normal

stress. The isotropic elastic stress strain law gives:

_ 6 u r _ o r oe+oz + 9( t) 87a,’ _E E O‘ r’

16%, u, 09 Or+Oz = — —— = — —— + s a" r as r E E O‘ (m)

(8)

_ 0 u z _ o z o r + o + 9( t) e,— 62 — E v or r,

where E is the modulus of elasticity, v the Poisson ratio and or is the linear coefficient of

thermal expansion.

Twoparticular stress strain cases are considered for the elastic region: (i) plane stress

where the normal stress (61) is equal to zero and (ii) plane strain where the strain (sZ) is

equal to zero. In both cases the material behavior outside the reverse cyclic plastic zone

is modeled by an isotropic thermo-elastic stress strain law. This set of equations is

reduced to the same form, for example for plane stress they are reduced to the first two

with 61 set equal to zero. For plane strain upon solving the last stress-strain equation for

62 and substituting it into the first two the same form is found with altered elastic and

thermal constants. The plane stress form will be adopted here for simplicity and the

plane strain alteration will only be noted in some final results. The plane stress case

gives:

@ u u 0 0'6

u o 0

’—’=—’—— +9 ,t d —’=—"— —’+ 9 ,t 9 6 r r E v E O L ( r ) a n r E \ / E aO( r )

Fromthe second of these, multiplying by r and differentiating, results in

5m 6

I" % — v % + o l 9 ( r , t ) )

0r : 5

(10)

Equating this to the first equation of the two above and rearranging, as well as

introducing the equilibrium equation to eliminate 66, gives:

52