A large steel plate under uniform applied tensile stress contains two cracks, bo

Question

A large steel plate under uniform applied tensile stress contains two cracks, both of whiclh are remote from the edges of the plate and from each other. One is 12 mm in length and is inclined at an angle = 30°, the other is 15 mm in length and is inclined at an angle = 45° , see figure below. The steel has fracture toughness of KJC-23 MPa/m and Calculate the fracture stress of the plate and state which of the two cracks will cause failure and in which failure mode. a) b) Using the maximum hoop stress criterion, determine the direction of the crack propagation for both cracks. C) r a

Explanation / Answer

1 Facture Stress of Plate & failure mode. Kc=15mpa(m)0.5 K=BX stress (3.14X Crack Length)^0.5 a 12 mm B 15 mm Kc=K=15= 1*Stress*(3.14*12 x10^-3)^0.5 Stress 77.3 Mpa The uniaxial stress that will cause yielding is given by We Note sy/Stress 240/77.3 3.1 yield stress 240 Mpa so,Facture Occurs before Yeilding. 2 Kc=15mpa(m)0.5 K=BX stress (3.14X Crack Length)^0.5 a 12 mm B 15 mm Kc=K=15= 1*Stress*(3.14*15 x10^-3)^0.5 Stress 69.12 Mpa The uniaxial stress that will cause yielding is given by We Note sy/Stress 240/69.12 3.47 yield stress 240 Mpa so, Facture Occurs before Yeilding. 3 HOOP STRESS CRITERION For a GC = 130 kJ/m2. Consider V, Poisson's ratio 0.3 E 210 GPa a=((E)/(3.14*Stress^2*(1-V^2))^0.5 a= ((130*10^3*210*10^9)/(3.14*(77.3*10^6)^2*(1-0.3^2)))^0.5 the critical crack size that caused failure a 1.2 mm For B GC = 130 kJ/m2. Consider V, Poisson's ratio 0.3 E 210 GPa a=((E)/(3.14*Stress^2*(1-V^2))^0.5 b= ((130*10^3*210*10^9)/(3.14*(69.12*10^6)^2*(1-0.3^2)))^0.5 the critical crack size that caused failure b 1.4 mm

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