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∓ESTIMATION OF PLASTIC ZONE SIZE IN LINEAR ELASTIC MATERIAL

Abstract: We know because of plastic zone created in the vicinity of crack, crack behaves as if it were slightly longer. As stress in continuously applied plastic region goes on increasing and when it becomes comparable to specimens dimensions, linear elastic stress analysis becomes increasingly inaccurate. So it becomes necessary to estimate plastic zone size, in order to decide which parameters to be applied in fracture and crack growth estimation.

   Here I propose of using von misses stress criteria as basic criteria for evaluation of plastic zone size.

Introduction Earlier approaches include THE IRWIN APPROACH and THE STRIP YEILD MODEL in estimation of plastic zone and both of them leads to simple corrections for crack tip yielding. THE IRWIN APPROACH includes normal uniaxial failure criteria in evaluating plastic zone size, whereas THE STRIP YEILD MODEL assumes a central crack with slender plastic zoneand closure stress is assumed to be σ_y applied at each crack tip. We all know, von misses stress criteria is,

                                              σ_e = 1/√2 √((σ_1- σ_2 )^2+(σ_2- σ_3 )^2+(σ_3- σ_1 )^2 )  .................1

Where, σ_e = effective stress

        σ_1 = 1st principal stress
       〖 σ〗_2 = 2nd principal stress
        σ_3 = 3rd principal stress

Now here we are estimating plastic zone size for mode l type of fracture, σ_(1,) σ_2 = (σ_xx+σ_yy)/2 ± √(((σ_xx-σ_yy)/2)^2+(τ_xy )^2 )...................................................................................2

      From stress intensity relations we can say, assuming plane stress condition, 

i.e. σ_zz = 0 and also τ_yz=0 τ_xz=0 Under the situation shown,

    σ_xx = K_1/√2πr  cos⁡〖θ/2〗 [1-sin⁡〖θ/2〗  sin⁡〖3θ/2〗] .......................3
    σ_yy = K_1/√2πr  cos⁡〖θ/2〗 [1+sin⁡〖θ/2〗  sin⁡〖3θ/2〗]......................  4 
    τ_xy =  K_1/√2πr  cos⁡〖θ/2〗  sin⁡〖θ/2〗  sin⁡〖3θ/2〗.................................5
Now putting back equations 3,4and 5 in equation 2 we get,

σ_(2 ,) σ_1= K_1/√2πr cos⁡〖θ/2〗 [1∓〖√2 sin〗⁡〖θ/2〗 sin⁡〖3θ/2〗] and σ_(3 )=0....................................................6 Substituting these values in equation 1.......we will have .......... σ_e= 1/√2 √((K_1/√2πr cos⁡〖θ/2〗 〖2√2 sin〗⁡〖θ/2〗 sin⁡〖3θ/2〗 )^2+(K_1/√2πr cos⁡〖θ/2〗 [1-〖√2 sin〗⁡〖θ/2〗 sin⁡〖3θ/2〗])^2+(K_1/√2πr cos⁡〖θ/2〗 [1+〖√2 sin〗⁡〖θ/2〗 sin⁡〖3θ/2〗])^2 ) σ_e= 1/√2 K_1/√2πr cos⁡〖θ/2〗 √(2+12 〖〖(sin〗⁡〖θ/2)〗〗^2 〖〖(sin〗⁡〖3θ/2)〗〗^2 ) For in plane crack we have, θ=0. σ_e= K_1/√2πr

σ_ys= K_1/√2πr

On solving further, we get

r= 1/2π (K_1/σ_ys )^2

Result: Result that we have obtained is also consistent with results obtained by irwin’s strip yield model that uses maximum normal stress theory.

Category:Fracture mechanics