Chap 08 Solns-6E - CHAPTER 8 FAILURE PROBLEM SOLUTIONS 8.1...

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CHAPTER 8 FAILURE PROBLEM SOLUTIONS 8.1 Several situations in which the possibility of failure is part of the design of a component or product are as follows: (1) the pull tab on the top of aluminum beverage cans; (2) aluminum utility/light poles that reside along freeways--a minimum of damage occurs to a vehicle when it collides with the pole; and (3) in some machinery components, shear pin are used to connect a gear or pulley to a shaft-- the pin is designed shear off before damage is done to either the shaft or gear in an overload situation. 8.2W The theoretical cohesive strength of a material is just E /10, where E is the modulus of elasticity. For the ceramic materials listed in Table 12.5, all we need do is divide E by 10, and therefore Si 3 N 4 --30.4 GPa (4.4 x 10 6 psi) ZrO 2 --20.5 GPa (3.0 x 10 6 psi) SiC--34.5 GPa (5.0 x 10 6 psi) Al 2 O 3 --39.3 GPa (5.7 x 10 6 psi) Glass ceramic--12.0 GPa (1.7 x 10 6 psi) Mullite--14.5 GPa (2.1 x 10 6 psi) MgAl 2 O 4 --26 GPa (3.8 x 10 6 psi) MgO--22.5 GPa (3.3 x 10 6 psi) Fused silica--7.3 GPa (1.1 x 10 6 psi) Soda-lime glass--6.9 GPa (1.0 x 10 6 psi ) 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 1
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= (2)(170 MPa) 2.5 x10 2 mm 2 4 mm 1/2 = 2404 MPa (354,000 psi) 8.4 In order to estimate the theoretical fracture strength of this material it is necessary to calculate σ m using Equation (8.1) given that σ o = 1035 MPa, a = 0.5 mm, and ρ t = 5 x 10 -3 mm. Thus, σ m =2 σ o a ρ t = (2)(1035 MPa) 0.5 mm 5x10 3 mm = 2.07 x 10 4 MPa = 207 GPa 3 x 10 6 psi ( ) 8.5 In order to determine whether or not this ceramic material will fail we must compute its theoretical fracture (or cohesive) strength; if the maximum strength at the tip of the most severe flaw is greater than this value then fracture will occur--if less than, then there will be no fracture. The theoretical fracture strength is just E /10 or 25 GPa (3.63 x 10 6 psi), inasmuch as E = 250 GPa (36.3 x 10 6 psi). The magnitude of the stress at the most severe flaw may be determined using Equation (8.1) as σ m σ o a ρ t = (2)(750 MPa) (0.20 mm) /2 0.001mm = 15 GPa 2.2 x 10 6 psi ( ) Therefore, fracture will not occur since this value is less than E /10. 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
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= (2) 393 x 10 9 N/m 2 () (0.90 N/m) ( π ) 4x10 4 m 2 = 33.6 x10 6 N/m 2 = 33.6 MPa 8.7 The maximum allowable surface crack length for MgO may be determined using Equation (8.3); taking the value of 225 GPa as the modulus of elasticity (Table 12.5), and solving for a , leads to a= 2E γ s πσ c 2 = (2) 225 x 10 9 2 ( ) (1.0 N/ m) ( π )13 .5 x10 6 2 2 = 7.9 x 10 -4 m = 0.79 mm (0.031 in.) 8.8W This problem calls for us to calculate the normal σ x and σ y stresses in front on a surface crack of length 2.0 mm at various positions when a tensile stress of 100 MPa is applied. Substitution for K = σ π a into Equations (8.9aW) and (8.9bW) leads to σ x = σ f x ( θ ) a 2r σ y = σ f
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This homework help was uploaded on 02/07/2008 for the course MATENG MAT201 taught by Professor Na during the Spring '08 term at Wisconsin Milwaukee.

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Chap 08 Solns-6E - CHAPTER 8 FAILURE PROBLEM SOLUTIONS 8.1...

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