MSE 2610 Mechanical Properties of Materials Fall 2009 Homework 8 Answers Problem 1: 25 points – 5 pts each, 3 for microstructure, 2 for percentages Using the Time-Temperature-Transformation (TTT) diagram shown below for a 0.45 wt% C steel alloy, specify the nature of the final microstructure (in terms of phases present and approximate percentages of each) of a small specimen that has been subjected to the following time-temperature treatments. In each case, assume that the specimen begins at 845°C, and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Rapidly cool to 250°C, hold for 103 s, then quench to room temperature. 100% martensite (or mostly martensite) (b) Rapidly cool to 450°C, hold for 10 s, then quench to room temperature. (±5% leeway) Bainite = Ferrite+ Cementite Lever Rule : 0.45 − 6.7~ 0.01− 6.7 ≈ 93.4% Ferrite, 6.6% Cementite 65% bainite, remainder [mostly] martensite barb2right ~ 61% Ferrite, 4% Cementite, 35% martensite (c) Rapidly cool to 625°C, hold at this temperature for 10 s, rapidly cool to 400°C, hold at this temperature for 5 s, then quench to room temperature. 100% Pearlite (same lever rule approximately) barb2right 93.4% Ferrite, 6.6% Cementite (d) Rapidly cool to 750°C, hold for 105 s, then quench to room temperature. 750°C is solutionizing temperature according to TTT, so it doesn’t matter how long we hold it there. It’s as if we quenched immediately, so 100% martensite (or mostly martensite). (e) Rapidly cool to 750°C, hold for 105 s, then rapidly cool to 625°C, hold at this temperature for 10 s, then rapidly cool to 400°C, hold at this temperature for 5 s, then quench to room temperature. Again, solutionize first (doesn’t change anything), then it’s the same as (c). Problem 2: 35 points The data were obtained for creep of a polycrystalline oxide having a grain size of 10 µm. Strain rate (s-1) Stress (MPa) T = 1750 K T = 1800 K 10 2.0 x 10-8 7.0 x 10-8 20 4.0 x 10-8 1.4 x 10-7 30 6.0 x 10-8 2.1 x 10-7 40 8.0 x 10-8 2.8 x 10-7 50 2.3 x 10-7 8.0 x 10-7 60 5.0 x 10-7 1.8 x 10-6 70 9.9 x 10-7 3.5 x 10-6 80 1.8 x 10-6 6.3 x 10-6 90 3.1 x 10-6 1.1 x 10-5 100 4.9 x 10-6 1.7 x 10-5 a) 25 points: 10 pts for nc, 10 pts for QC, 5 pts for showing how you got those values (equation analysis and/or plot). Determine the stress exponent, nC, and activation energy QC. (This will probably involve some sort of plot) HW 8, Problem 2 y = 4.4714x - 31.555 y = x - 18.777 y = 4.4718x - 32.815 y = x - 20.03 -20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 2 2.5 3 3.5 4 4.5 5 Ln(Stress) 1750 Low Stress 1750 High Stress 1800 Low Stress 1800 High Stress Linear (1800 High Stress) Linear (1800 Low Stress) Linear (1750 High Stress) Linear (1750 Low Stress) Now we analyze this data, given the following generic creep equation: &εSS = Aσ n exp −∆QkT barb2right natural log, ln &εSS( )= ln A( )+ nlnσ − ∆QkT From this we see that the stress exponent can be read from the slope of a stress-strain rate graph. As such, we see that at low stress the stress exponent is n = 1.0 and for high stress the stress exponent is n = 4.47. (We must have continuous creep, so we include the 4th data point in both low-stress and high-stress lines). This is consistent with the high-stress regime being dominated by power law (P-L) creep and the low-stress regime dominated by either Coble or Nabarro-Herring creep. We now seek to determine the value of ∆Q for the high-stress and low-stress regimes. We may simply subtract the values of the logarithm of the strain rate at some stress for both the high-stress and low stress regimes to get the following. ( )[ ] −∆−=∆ 12 11ln TTk Q SSε& barb2right ( )[ ] 21 11 ln TT kQ SS − ∆=∆ ε& Looking at the spreadsheet values for change in the logarithm of the strain rate we see that the value is constant for both low-stress and high-stress regimes. Since T1 and T2 are also constant along with k, we conclude that delta Q is constant for both stress regimes. Substituting values gives, ∆Q = 6.82 eV/atom. (averaged over all strain rates) b) 10 points: 3 points for answer, 3 points for noting same Qc for both mechanisms, 4 points for noting that NH and PLC are both limited by vacancy diffusion, and Coble isn’t. Does the diffusional creep region correspond to Nabarro-Herring creep or to Coble creep? Hint: Remember how each type depends on activation energy. Power law creep moves by dislocation motion (limited by the diffusion of vacancies). Nabarro-Herring creep moves by atomic diffusion through the lattice (limited by the diffusion of vacancies). Coble creep moves by atomic diffusion around grain boundaries (limited by number of atoms and circuitous pathways). We see both PLC and NH creep are limited by the same mechanisms, so now we look at our data. We see the same Qc for both high-stress and low-stress creep, which indicates that both are limited by the same mechanism. Therefore, the diffusional creep must be NH creep (since Coble has a lower Qc than NH). If the Qc were different, it would be Coble creep. Problem 3: 20 points You are to make a widget that can sustain a maximum strain of 0.1. You want to make it out of pure nickel with a grain size of 1mm. The deformation map is given below (make sure it’s the nickel map, and not copper). Note the thin lines marked as powers of 10 indicate lines of constant strain rate (in units of s-1). (a) 6 points: 3 for strain rate calculation, 3 for homologous temperature If the part must last for 10 years operated at a normalized shear stress (σs/µ) of 10-4, what is the maximum homologous temperature (T/TM) at which it can be used? &ε ⋅t = ε → &ε = εt = 0.110yrs⋅365days⋅24hrs⋅3600s ≈ 3E −10s−1 Using the strain rate and the normalized shear stress, we use the map to find the homologous temperature is ~0.5. (b) 4 points: At the temperature you’ve found in part a, what is the dominant mechanism of deformation? Directly from the map, we see the dominant mechanism (at 0.5 T/TM and 10-4 normalized shear stress) is PLC. (c) 10 points: At the temperature from part a, how does the time to failure and the dominant deformation mechanism change if: i. 5pts: the applied stress is decreased by an order of magnitude? The mechanism changes (3 pts) to diffusional creep, and time to failure (2 pts) increases by large amount (strain rate decreases past where we can measure) ii. 5 pts: the allowable strain is increased by a factor of 4? Strain rate does not change, so mechanism does not change (3 pts), so time to failure increases by factor of 4 (2 pts – 1 pt if just “increases”). Problem 4: 20 points A pure copper bar 1 m long of rectangular cross section (2.5 cm x 4 cm) is used to support a load of 200 kg. The bar is held at a constant temperature of 770 °C. (shear strength) µCu = 46 GPa a) What is the steady-state creep rate b) What is the dominant creep mechanism at this point? c) If failure occurs when the part increases its length by 3 mm, how long will the part last? d) What might be done to the bar to increase its service lifetime? Note: Use the deformation mechanism map on the following page to answer this question. Solution: Area of cross-section = 10-3 m2 Load = 1962 N barb2right σnormal = 1.962 MPa τmax = σnormal/2 = 0.981 MPa (some people read graph as σnormal / µ, so accept that due to blurry image) a) 5 pts – 3 pts for normalized stress, 2 pts for reading graph: In order to read off the steady-state creep rate we need to normalize the shear stress by the shear modulus of copper (µ = 46 GPa). σS µ = 0.981MPa 46000MPa = 2.13×10 −5 or σN µ = 4.26 ×10 −5 Reading this value on the y-axis with 740°C from the x-axis give a steady-state creep rate of approximately 1-5x10-8 s-1. b) 5 pts: Will accept PLC or NH c) 5 pts – 2 for strain, 3 for time: We can give an upper bound on the lifetime based on SS t εε&≤ . A length increase in 3 mm gives a strain of ε = 0.003. This number does not significantly change if we consider true strain. So we can say that t ≅ 1000-5000 min. d) 5 points – must name one valid reason: (partial list) If NH, we can increase the grain size, since 21dSS ∝ε& . For either PLC or NH, we can increase cross-sectional area, or change to stronger material (make µ larger). Drew Microsoft Word - HW 8 solutions.doc