Crack Tip Plastic Zone According to Irwin’s Model

Abstract

Consider a central crack of length 2a in an infinite plate subjected to uniaxial stress σ at infinity perpendicular to the crack plane. According to the Irwin model, the effective crack is larger than the actual crack by the length of plastic zone. Show that the stress intensity factor corresponding to the effective crack, called effective stress intensity factor K_eff, for conditions of plane stress, is given by

EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4samaaBa % aaleaacaWGLbGaamOzaiaadAgaaeqaaOGaeyypa0ZaaSaaaeaacqaH % dpWCdaqadaqaaiabec8aWjaadggaaiaawIcacaGLPaaadaahaaWcbe % qaamaalyaabaGaaGymaaqaaiaaikdaaaaaaaGcbaWaamWaaeaacaaI % XaGaeyOeI0IaaGimaiaac6cacaaI1aWaaeWaaeaadaWcgaqaaiabeo % 8aZbqaaiabeo8aZnaaBaaaleaacaWGzbaabeaaaaaakiaawIcacaGL % PaaadaahaaWcbeqaaiaaikdaaaaakiaawUfacaGLDbaadaahaaWcbe % qaamaalyaabaGaaGymaaqaaiaaikdaaaaaaaaaaaa!512E!]]</EquationSource><EquationSource Format="TEX"><![CDATA[$$ {K_{eff}} = \frac{{\sigma {{\left( {\pi a} \right)}^{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-\nulldelimiterspace} 2}}}}}{{{{\left[ {1 - 0.5{{\left( {{\sigma \mathord{\left/ {\vphantom {\sigma {{\sigma _Y}}}} \right. \kern-\nulldelimiterspace} {{\sigma _Y}}}} \right)}^2}} \right]}^{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-\nulldelimiterspace} 2}}}}} $$

(1)

Then, consider a large plate of steel that contains a crack of length 20 mm and is subjected to a stress σ = 500 MPa normal to the crack plane. Plot the σ_y stress distribution directly ahead of the crack according to the Irwin model. The yield stress of the material is 2000 MPa.