HW10 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK 10 FAILURE (Chapter 8) PROBLEM SOLUTIONS Principles of Fracture Mechanics 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 =2 0 a t 1/2 = (2)(170 MPa) 2.5 10 2 mm 2 2.5 10 4 mm 1/2 = 2404 MPa (354,000 psi)
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8.2 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 (0.01 in.) and having a tip radius of curvature of 1.2 10 -3 mm (4.7 10 -5 in.) when a stress of 1200 MPa (174,000 psi) 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 =2 0 a t 1/2 = (2)(1200 MPa) 0.25 mm 1.2 10 3 mm 1/2 = 3.5 10 4 MPa = 35 GPa (5.1 10 6 psi )
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8.3 If the specific surface energy for soda-lime glass is 0.30 J/m 2 , using data contained in Table 12.5, compute the critical stress required for the propagation of a surface crack of length 0.05 mm. Solution We may determine the critical stress required for the propagation of an surface crack in soda-lime glass using Equation 8.3; taking the value of 69 GPa (Table 12.5) as the modulus of elasticity, we get c = 2 E s a 1/2 = (2) ( 69 10 9 N/m 2 ) (0.30 N/m) ( )0 .05 10 3 m  = 16.2 10 6 N/m 2 = 16.2 MPa
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8.4 A polystyrene component must not fail when a tensile stress of 1.25 MPa (180 psi) is applied. Determine the maximum allowable surface crack length if the surface energy of polystyrene is 0.50 J/m 2 (2.86 10 -3 in.-lb f /in. 2 ). Assume a modulus of elasticity of 3.0 GPa (0.435 10 6 psi). Solution The maximum allowable surface crack length for polystyrene may be determined using Equation 8.3; taking 3.0 GPa as the modulus of elasticity, and solving for a , leads to a = 2 E s  c 2 = (2) ( 3 10 9 N/m 2 ) (0.50 N/m) ( )(1.25 10 6 N/m 2 ) 2 = 6.1 10 -4 m = 0.61 mm (0.024 in.)
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8.5 A specimen of a 4340 steel alloy having a plane strain fracture toughness of 45 MPa m ( 41 ksi in . ) is exposed to a stress of 1000 MPa (145,000 psi). Will this specimen experience fracture if it is known that the largest surface crack is 0.75 mm (0.03 in.) 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 K Ic , Y , and the largest value of a in the material. This requires that we solve for c from Equation 8.6. Thus c = K Ic Y a = 45 MPa m (1.0 ) ( ) ( 0.75 10 3 m ) = 927 MPa (133,500 psi)
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This note was uploaded on 09/11/2011 for the course ENG 407 taught by Professor Tsakal during the Spring '11 term at Rutgers.

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HW10 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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