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Unformatted text preview: ENGR 4250U Winter 2011 Final Exam Review Problems Final Exam Review: Ch 8, 1421 excluding 20 Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of Chapter 8 35 MPa m (31.9 ksi in. ). It has been determined that fracture results at a stress of 250 MPa (36,250 psi) when the maximum (or critical) internal crack length is 2.0 mm (0.08 in.). For this same component and alloy, will fracture occur at a stress level of 325 MPa (47,125 psi) when the maximum internal crack length is 1.0 mm (0.04 in.)? Why or why not? We are asked to determine if an aircraft component will fracture for a given fracture toughness (35 Solution MPa m ), stress level (325 MPa), and maximum internal crack length (1.0 mm), given that fracture occurs for the same component using the same alloy for another stress level and internal crack length. It first becomes necessary to solve for the parameter Y , using Equation 8.5, for the conditions under which fracture occurred (i.e., = 250 MPa and 2 a = 2.0 mm). Therefore, Y = K Ic a = 35 MPa m (250 MPa) ( ) 2 10 3 m 2 = 2.50 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 K Ic for the alloy. Thus, Y a = (2.50)(325 MPa) ( ) 1 10 3 m 2 = 32.2 MPa m (29.5 ksi in. ) Therefore, fracture will not occur since this value ( 32.3 MPa m ) is less than the K Ic of the material, 35 MPa m . ENGR 4250U Winter 2011 Final Exam Review Problems The density and associated percent crystallinity for two polytetrafluoroethylene materials are as follows: Chapter 14 ( g/cm 3 crystallinity ( % ) ) 2.144 51.3 2.215 74.2 (a ) Compute the densities of totally crystalline and totally amorphous polytetrafluoroethylene. (b ) Determine the percent crystallinity of a specimen having a density of 2.26 g/cm 3 . (a) We are asked to compute the densities of totally crystalline and totally amorphous polytetrafluoroethylene ( c and a from Equation 14.8). From Equation 14.8 let Solution C = % crystallinity 100 , such that C = c ( s a ) s ( c a ) Rearrangement of this expression leads to c ( C s s ) + c a C s a = 0 in which c and a are the variables for which solutions are to be found. Since two values of...
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 Spring '08
 Brenan

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