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Unformatted text preview: CHAPTER 8 FAILURE PROBLEM SOLUTIONS 8.3 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 σ m = 2 σ o a ρ t 1/ 2 = (2)(170 MPa) 2.5 x 10 2 mm 2 2.5 x 10 4 mm 1/2 = 2404 MPa (354,000 psi) 8.6 We may determine the critical stress required for the propagation of an internal crack in aluminum oxide using Equation (8.3); taking the value of 393 GPa (Table 12.5) as the modulus of elasticity, we get σ c = 2E γ s π a = (2) 393 x 10 9 N/ m 2 ( 29 (0.90 N/ m) ( π ) 4 x 10 4 m 2 = 33.6 x10 6 N/m 2 = 33.6 MPa 8.19 This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus (3.0 mm), the value of K Ic (98.9 MPa m ) , the design stress ( σ y /2 in which σ y = 860 MPa), and Y = 1.0. We first need to compute the value of a c using Equation (8.7); thus 196 a c = 1 π K Ic Y σ 2 = 1 π 98.9 MPa m (1.0 ) 860 MPa 2 2 = 0.0168 m= 16.8 mm (0.66 in.) Therefore, the critical flaw is subject to detection since this value of a c is greater than the 3.0 mm resolution limit. 8.28 This problem asks that we compute the maximum and minimum loads to which a 15.2 mm (0.60 in.) diameter 2014T6 aluminum alloy specimen may be subjected in order to yield a fatigue life of 1.0 x 10 8 cycles; Figure 8.44 is to be used assuming that data were taken for a mean stress of 35 MPa (5,000 psi). Upon consultation of Figure 8.44, a fatigue life of 1.0 x 10 8 cycles corresponds to a stress amplitude of 140 MPa (20,000 psi). Or, from Equation (8.16) σ max σ min = 2 σ a = (2)(140 MPa) = 280 MPa (40,000 psi) Since σ m = 35 MPa, then from Equation (8.14) σ max + σ min = 2 σ m = (2)(35 MPa) = 70 MPa (10,000 psi) Simultaneous solution of these two expressions for σ max and σ min yields σ max = +175 MPa (+25,000 psi) and σ min = 105 MPa (15,000 psi). Now, inasmuch as σ = F A o [Equation (6.1)], and A o = π d o 2 2 then F max = σ max π d o 2 4 = 175 x 10 6 N/ m 2 ( 29 ( π ) 15.2 x 10 3 m ( 29 2 4 = 31,750 N (7070 lb f ) F min = σ min π d o 2 4 = 105 x 10 6 N / m 2 ( 29 ( π ) 15.2 x 10 3 m ( 29 2 4 = 19,000 N ( 4240 lb f ) 8.48 (a) We are asked to estimate the activation energy for creep for the low carbonnickel alloy having the steadystate creep behavior shown in Figure 8.29, using data taken at σ = 55 MPa (8000 psi) and temperatures of 427 ° C and 538 ° C. Since σ is a constant, Equation (8.20) takes the form 197 Ý ε s = K 2 σ n exp  Q c RT = K 2 ' exp  Q c RT where K 2 ' is now a constant. (Note: the exponent n has about the same value at these two temperatures per Problem 8.47.) temperatures per Problem 8....
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This note was uploaded on 05/03/2010 for the course ME 250750 taught by Professor Signer during the Summer '10 term at Wichita State.
 Summer '10
 Signer
 Stress

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