Practice Problems on Fracture for the Mid term Determine what equations apply what the variables are in the equations and what they mean and solve for the requested information What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2 5 10 4 mm and a crack length of 2 5 10 2 mm when a tensile stress of 170 MPa is applied 8 1 What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2 5 10 4 mm 10 5 in and a crack length of 2 5 10 2 mm 10 3 in when a tensile stress of 170 MPa 25 000 psi is applied Solution This problem asks that we compute the magnitude of the maximum stress that exists at the tip of an internal crack Equation 8 1 is employed to solve this problem as m a t Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0 25 mm and having a tip radius of curvature of 1 2 10 3 mm when a stress of 1200 MPa is applied Solution In order to estimate the theoretical fracture strength of this material it is necessary to calculate m using Equation 8 1 given that 0 1200 MPa a 0 25 mm and t 1 2 10 3 mm Thus m a t A specimen of a 4340 steel alloy having a plane strain fracture toughness of is exposed to a stress of 1000 MPa Will this specimen experience fracture if it is known that the largest surface crack is 0 75 mm long Why or why not Assume that the parameter Y has a value of 1 0 Solution This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1000 MPa given the values of KIc Y and the largest value of a in the material This requires that we solve for c from Equation 8 6 Thus c Y K Ic a Therefore fracture will most likely occur because this specimen will tolerate a stress of 927 MPa 133 500 psi before fracture which is less than the applied stress of 1000 MPa 145 000 psi Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of It has been determined that fracture results at a stress of 250 MPa when the maximum or critical internal crack length is 2 0 mm For this same component and alloy will fracture occur at a stress level of 325 MPa when the maximum internal crack length is 1 0 mm Why or why not Solution MPa m Y K Ic a Now we will solve for the product Y a for the other set of conditions so as to ascertain whether or not this value is greater than the KIc for the alloy Thus Y a Therefore fracture will not occur since this value is less than the KIc of the material Calculate the maximum internal crack length allowable for a 7075 T651 aluminum alloy component that is loaded to a stress one half of its yield strength Assume that the value of Y is 1 35 for the geometry and K Ic 24 MPa m y 495 MPa for the alloy Solution This problem asks us to calculate the maximum internal crack length allowable for the 7075 T651 aluminum alloy in Table 8 1 given that it is loaded to a stress level equal to one half of its yield strength For this alloy KIc also y 2 495 MPa 2 248 MPa 36 000 psi Now solving for 2ac using Equation 8 7 yields ac K Ic Y
View Full Document
Unlocking...