113_2009_MT2_solns - MSE C113 Mechanical Behavior of...

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Unformatted text preview: MSE C113 Mechanical Behavior of Materials Professor Ritchie Midterm Exam 2 Solutions Problem 1 (35 Points) a) Creep Deformation mechanisms: Dislocation glide → barrier surmounted by thermal activation Dislocation creep → barrier surmounted by diffusional processes Diffusional creep → Coble and Nabarro‐Herring (boundary vs bulk diffusion) Grain Boundary Sliding Obstacles which inhibit dislocation movement will also reduce the creep rate at a given temperature. Thus precipitation/dispersion hardening, and solid solution strengthening will have an effect on creep properties. For creep purposes, grain size effects preclude use of grain boundary hardening. At low grain sizes, grain sliding is enhanced. Additionally, small grain sizes increase grain boundary diffusion rates in the Coble creep regime. A good creep resistant alloy will have the following properties: Solid‐solution hardening Low diffusional rates Small stable precipitates or dispersoids High melting point Large grain size Good oxidation resistance i) Stainless steel → higher melting point ii) Precipitate → if the precipitates are stable, otherwise choose solid solution iii) Coarse grained → both grain boundary and bulk diffusion rates increase with decreasing grain size. In addition, grain sliding increases with decreasing grain size. b) i) Activation energy ඃ = "− ඃ Δ ln⁡ඃඃඃ /ඃ0 ) (̇ ̇ + Δ (1⁄ඃ) ඃඃඃඃඃඃඃඃ ඃ 11 Assume ඃ0 = 1 ̇ See figure below for strain rate – 1/T relationships: σ11 (psi) ̇ ss 1/T (R‐1) H (in‐lb) ‐3 ‐4 6000 10 6.35 x 10 8.93 x 10‐18 10‐5 6.7 x 10‐4 ‐3 ‐4 10000 10 6.65 x 10 6.25 x 10‐18 10‐5 7.15 x 10‐4 ‐3 ‐4 20000 10 7.2 x 10 6.25 x 10‐18 10‐5 7.7 x 10‐4 Note 1: Since there is a change in activation energy, this implies a change of mechanism. You cannot assume the same mechanism works for all three stress conditions. However, since we are interested in σ11 = 12000 psi, we can assume that the same mechanism is at work as for the 10ksi and 20ksi lines above. Note 2: the values of H may vary slightly depending on how you read the graph ii) Creep exponent. Δ ඃඃඃ10 (ඃඃඃ ) ̇ ඃ = " 7 Δ ඃඃඃ10 (ඃ11 ) ඃඃඃඃඃඃඃඃ ඃ There is a convenient line drawn at 1/T = 7 x 10‐4 R‐1. Reading off the graph, we have the following: σ11 (psi) ̇ ss m calculation ‐7 6000 ~3 x 10 log(2E‐5/3E‐7) /log(10000/6000) ‐5 10000 ~2 x 10 log(8E‐3/2E‐5) /log(20000/10000) ‐3 20000 ~8 x 10 log(8E‐3/3E‐7) /log(20000/6000) These points yield an average m = 8.44 iii) σ0 can be calculated simply using the following expression: 1 1/ඃ − ඃ ඃ0 = (ඃ11 ) 8 : ඃ ඃඃඃ ඃඃ̇ ඃ m 8.22 8.64 8.46 σ11 = 12 ksi Find ̇ ss at this stress: solve the m equation: 10 ^ [m*log(12 ksi / 10 ksi)] =( ̇ ss at 12 ksi) / (̇ ss at 10 ksi) 10 ^ [8.44 *log(12 / 10)] =( ̇ ss at 12 ksi) / (2 x 10‐5) ̇ ss at 12 ksi = 9.3 x 10‐5 1/8.44 6.25 ×10 −18 1 − ඃ0 = (12000 ඃඃඃ ) 8 : ඃ 8.44 ×6.79×10 −23 ×1428 9.3 × 10−5 σ0 = 17.4 psi. iv) Elastic creep of interest el = σ11 / E = 12000 psi / 30 x 106 psi = 4 x 10‐4. This is a lot smaller than 0.10, so we can ignore the elastic portion of creep. We are told to assume primary creep is negligible. ̇ 11 = constant and 11 = 0.1 t = 11 /̇ 11 Substituting into the dimensionless creep equation: let ඃ0 = 1 ̇ ඃ11 ඃ − ඃ ඃ11 = 8 : ඃ ඃඃ ̇ ඃ0 12000 8.44 = > 17 .4 ? −6.25 ×10 −18 ඃඃඃ >6.79×10 −23 ×1460 ? = 3.78 × 10−4 t = 0.1 / 0.0004 = 264.5 hours. Problem 2 (40 Points) Problem 3 (40 Points) a) b) c) i) ii) iii) 2 2 1 ඃඃඃ 1 30 ඃඃඃ √ඃ 15ඃඃ = D G= D G = 0.002381 ඃ 2ඃ ඃඃ 2ඃ 950 ඃඃඃ ඃ = 2.5 ඃඃ = 0.0025 ඃ 15ඃඃ < ඃ ∴ ඃඃ ඃඃඃඃඃ ඃඃඃඃඃඃ ඃඃ = 2.73ඃඃඃ Nඃ 50 × 10−6 = 32.5 ඃඃඃ √ඃ ඃඃ > ඃඃඃ ∴ ඃඃඃඃ ඃඃඃඃඃඃඃඃ ඃඃඃඃ ඃඃඃඃඃඃඃඃඃඃඃඃඃඃ ...
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This note was uploaded on 01/21/2012 for the course MSE 113 taught by Professor Ritchie during the Fall '09 term at University of California, Berkeley.

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