Homework 4 on Dislocations, Yield Stress, Hardness, Creep, Grain Size 27-301, A. D. Rollett, Fall 2002 Chemical processing plant sometimes uses nickel or nickel-based alloys because of their corrosion resistance. It is important nonetheless to understand their strength in order to select the correct microstructural state. Taking the uncrystallized Nickel of homework 1 as our example, we are going to estimate its strength and suitability for service. 1a. [10 points] Based on a dislocation density of 10 14 m -2, estimate the mean spacing between dislocations. Explain how you made your estimate and, if you can, relate it to what has been discussed on stereology in class. Notes: you should first notice that dislocation density can be defined in two ways. The first is to consider the density as an amount of dislocation line length per unit volume (with units m/m 3, = m -2 ). You need to relate this to a density of points where dislocations intersect with a slip plane (with units 1/m 2 = m -2, again). If you consider dislocations arranged randomly in the material (which is the picture in Porter & Easterling) you may apply the standard stereological relationships that I gave you, table 2.1 from Underwood in lecture 1). If you consider a set of parallel dislocations, however, which is often done in textbooks for simplicity, and work it out directly, you will obtain an answer that is different by a factor of two. Do not be alarmed by this! This is a nice illustration of the importance of stereology. Finally, once you know the density of points that intersect a slip plane, you can estimate the spacing between them based, for example, on the area per dislocation (think of arranging the dislocation intersection points on a regular lattice). There is also a stereological relationship between the density of points on a plane (P A ) and the mean spacing between those points; we did not have time to go into this in class. I will make some supplemental slides available on this topic. Answer: For randomly arranged dislocations, the relationship P A =L V /2 applies. For straight and parallel dislocations, P A =L V applies since all the dislocations thread through the slip plane. Then we need an estimate of the mean spacing. Not discussed in class is the case for random dislocations, for which 2 =(2 P A ) -1 ; for straight and parallel dislocations, 2 =( P A ) -1. So for the former case (random), l=(2 {L V /2}) -1 and for the latter case (straight), l=1/ L V. Based on L V =10 14.m -2, l= 71 nm (random) or 100 nm (straight). 1b. [10 points] Now estimate the critical resolved shear stress of the nickel based on the relations given in class, given that the shear modulus is 76 GPa and the Burgers vector is 0.25 nm. Using t= Gb/l, and assuming that we have straight and parallel dislocations, t = 190 MPa. Alternate answer #1:Using t= Gb/l, and assuming that we have randomly oriented dislocations, t = 271 MPa.
Alternate answer #2: use t= agb r, which gives t= 95 MPa. Note the difference in result! 1c. [10 points] What effect will raising the temperature have on this estimate of the flow stress? Hint: consider what happens to the shear modulus. The flow stress will tend to decrease with increasing temperature because the shear modulus decreases with temperature (in almost all materials). Thermal activation of dislocation motion (both conservative and non-conservative) will also lower the flow stress. Either of these is an acceptable answer. 1d. [20 points] In a single crystal tensile test, the orientation is given as [259]//tensile axis. Calculate the tensile yield stress based on the critical resolved shear stress that you obtained in part b. Note: Ni is fcc and can slip on any {111} plane in any <110> direction. You will have to find out which is the most highly stressed slip system, i.e. find the largest Schmid factor. This value of the Schmid factor is what you should use to determine the tensile yield stress because it determines which slip system will be activated first. The direction [259] is approximately in the center of the standard stereographic triangle (i.e. 001-011-111). A suggestion for how to proceed is to use a spreadsheet (e.g. excel) and make a list of all possible combinations of slip plane (111, -111, 1-11, -1-11) and slip direction (e.g. 111 is orthogonal to 110, -101 and 0-11), taking only positive versions of each (unit) vector. This will give you a table with 12 rows, one for each slip system (and 2 columns). Then calculate the dot products of the tensile axis with each pair of plane+direction in turn in order to obtain cosf and cosl, respectively (2 more columns). Then you can calculate the Schmid factor as cosf*cosl (1 more column). Finally, identify the row with the largest absolute value of the Schmid factor in it (i.e. positive or negative). {You can expand the table to include the negatives of each slip direction in addition: this will give you 24 rows (e.g. 111 is orthogonal to 110, -101, 0-11, 1-10, 10-1 and 01-1). If you go with the 24 row version, you will find that you obtain a pair of positive and negative Schmid factors for each pair of positive and negative slip directions. This positive/negative pairing corresponds to positive and negative directions of slip.} Answer: The largest Schmid factor is 0.49. If you used only 12 slip systems (no negatives) then you would have had to take the largest absolute value of the Schmid factor. Negative Schmid factors simply mean slip in the opposite direction on the same slip system. If you used 24 systems, then you can take the largest positive value. Based on this analysis, the tensile yield stress is 190/0.49, or, s = 388 MPa. [20] 1e. [15 points] The orientation of several grains has been characterized in a polycrystal: their orientations with respect to the tensile axis are [259], [001], [011] and [111]. Calculate their Schmid factors and calculate their average in order to estimate the yield stress of the polycrystal. You may find it helpful to draw a stereographic projection and note the locations of the slip direction and plane on it that satisfy the maximum Schmid
factor criterion. Obviously the symmetry axes give an ambiguous result in that several slip systems satisfy the criterion with equal Schmid factors. [Your answers should be close to 0.49, 0.43, 0.425, and 0.275, respectively; you should be able to use the same procedure to solve the problem as in part (e) above]. Answer: the average Schmid factor = (0.49+0.43+0.425+0.275)/4=0.405. Therefore the polycrystal should yield at 190/0.405 = 469 MPa. The other Schmid factors can be calculated by replacing [259] as the tensile axis direction with the other three directions. Note that these more symmetric directions give more than one system with the same factor, indicating that more than one slip system is active even when only single slip is considered. 1f. [10 points] Unfortunately the above estimate is not very accurate because multiple slip is necessary for a polycrystal to yield. The Taylor factors of the above orientations are 3.010, 2.449, 3.674, and 3.674, respectively. Compute the yield strength again from the average of the Taylor factors, and compare to the previous result. [In the graduate course, we would use a computer code to compute the Taylor factors, after having computed one by hand as an example.] Answer: the average Taylor factor, <M> = 3.20; therefore the yield stress would be 608 MPa. 1g. [10 points] If you were to measure the hardness of this nickel, what result would you expect in terms of a Vickers hardness (as you might obtain from a microhardness test)? A reasonable rule-of-thumb is that the hardness is approximately 3 times the yield stress. Thus H V ~ 3 x 608= 1825 MPa! This is admittedly a very high hardness and unlikely to be observed in practice. 1h. [5 points] Finally, what is your opinion of the suitability of this material for use in structural applications? Would you use this, or a nickel alloy that had been strengthened through some other means? Why? Answer: on the whole, this particular batch of nickel is a bit too hard to be useful. It would not fail gracefully if strained in tension, i.e. it would behave in a somewhat brittle manner. It would be likely to recrystallize if heated and so could not be exposed to elevated temperatures without losing strength. [5] 2a. [40 points] Nickel has a Hall-Petch coefficient of 0.13 MPa(m 3/2 ) and a friction stress of 60 MPa. Its shear modulus is 76 GPa and its Burgers vector is 0.25 nm. Assuming that the grain size (diameter) determines a minimum Orowan bowing stress (ignoring other dislocations that may be present within the grains), plot both the flow stress determined from the Hall-Petch equation, and the Orowan bowing stress (on the same
graph) as a function of grain size, d. Use logarithmic axes in order to cover the range 1nm d 1mm. Assume an average Taylor factor appropriate for randomly oriented polycrystals, <M>=3.1. Below what grain size does the Orowan bowing stress become dominant? Answer: using stresses calculated in the standard manner (see formulae on graph), the Orowan stress becomes dominant below about 140 nm. Note the Taylor factor is needed to convert a critical resolved shear stress based on the Orowan dowing stress to a yield stress (whereas the Hall-Petch formula is already in terms of macroscopic yield stress). Flow Stress (Pa) 10 11 10 10 10 9 10 8 10 7 10 6 10 5 creep.strength.hwk4.f02.kdata Orowan stress = <M>Gb/l Hall-Petch = s friction +kd -1/2 Cross-over grain size = 144 nm 10 4 1 10 100 1000 10 4 10 5 10 6 Grain Size (nm) 2b. [30 points] Explaining the flow stress of very fine-grained materials is challenging. Based, however, on what you have learned about creep, it is useful to calculate the expected creep rates based on diffusion mechanisms. Use the following equations and data for Nabarro-Herring and Coble creep at room temperature (300K) in Cu. Nabarro- Herring creep is the bulk diffusion mechanism discussed in class for creep, and Coble creep is the equivalent diffusion-based creep mechanism except that the diffusion takes place along grain boundaries (instead of the bulk). Nabarro-Herring Creep:
Ê D e = A bulk ˆ Ê NH Á sw ˆ Á Ë d 2 Ë kt Coble Creep: Ê D e = A gb d ˆ Ê Coble Á Ë d 3 sw ˆ Á Ë kt For bulk diffusion, Q bulk = 200,000 J/mole, D 0 = 31 mm 2 /s (be careful of units) and A NH = 10. For grain boundary diffusion, Q gb = 100,000 J/mole, D 0 = 50 mm 2 /s, d = 0.25nm, A Coble = 40. Plot the creep rate for both mechanisms against grain size, d, over the same range of grain size and flow stress as calculated in part (a); use the larger of the two flow stresses at each grain size. You should find that the creep rate (in units of "per second") becomes appreciable at small enough grain sizes for one of the mechanisms despite the low temperature. Based on this result, comment on what this means for the chances of observing high strength in nanocrystalline (pure) copper. Try repeating your calculations (in a spreadsheet, I trust!) for 400K. What hope is there for high strength at (slightly) elevated temperature?! Nabarro-Herring, room T Coble, room T NH (400K) Coble (400K) 1000 1 0.001 10-6 Creep Rate 10-9 10-12 10-15 10-18 10-21 10-24 10-27 10-30 10-33 10-36 1 10 100 1000 10 4 10 5 10 6 Grain Size (nm)
Clearly, at small enough grain sizes, the creep rate (in Coble creep, but not N-H creep) becomes appreciable for stresses close to the nominal yield stress. Raising the temperature only makes things worse. Moreover one could reasonably expect grain growth to be very rapid for nanocrystalline materials since the driving force rises to significant levels for nanocrystalline materials. Copper is also prone to dynamic recrystallization at temperatures not far above room temperature which would also tend to coarsen the microstructure. Generally, the only hope for high strengths from nanocrystalline materials is to stabilize the microstructure with second phase particles. One final comment is that it is, obviously, a bit artificial to take stress values from a different material, but nickel is very similar to copper and the creep rate is more sensitive to grain size than to stress, so the approach is reasonable.