Homework 4 on Dislocations, Yield Stress, Hardness, Creep, Grain Size27-301, A. D. Rollett, Fall 2002Chemical processing plant sometimes uses nickel or nickel-based alloys because of theircorrosion resistance. It is important nonetheless to understand their strength in order toselect the correct microstructural state. Taking the uncrystallized Nickel of homework 1as our example, we are going to estimate its strength and suitability for service.1a. [10 points] Based on a dislocation density of 1014m-2, estimate the mean spacingbetween dislocations. Explain how you made your estimate and, if you can, relate it towhat 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 unitvolume (with units m/m3, = m-2). You need to relate this to a density of pointswhere dislocations intersect with a slip plane (with units 1/m2 = m-2, again). Ifyou consider dislocations arranged randomly in the material (which is the picturein Porter & Easterling) you may apply the standard stereological relationships thatI gave you, table 2.1 from Underwood in lecture 1). If you consider a set ofparallel 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 oftwo. Do not be alarmed by this! This is a nice illustration of the importance ofstereology. Finally, once you know the density of points that intersect a slipplane, you can estimate the spacing between them based, for example, on the areaper dislocation (think of arranging the dislocation intersection points on a regularlattice). There is also a stereological relationship between the density of points ona plane (PA) and the mean spacing between those points; we did not have time togo into this in class. I will make some supplemental slides available on this topic.Answer: For randomly arranged dislocations, the relationship PA=LV/2 applies.For straight and parallel dislocations, PA=LV applies since all the dislocationsthread 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√PA)-1;for straight and parallel dislocations, ∆2=(√PA)-1. So for the former case (random),l=(2√{LV/2})-1and for the latter case (straight), l=1/√LV. Based on LV=1014.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 therelations given in class, given that the shear modulus is 76 GPa and the Burgers vector is0.25 nm.Using t= Gb/l, and assuming that we have straight and parallel dislocations, t = 190MPa.Alternate answer #1:Using t= Gb/l, and assuming that we have randomly orienteddislocations, t = 271 MPa.Alternate answer #2: use t= aGb√r, which gives t= 95 MPa. Note the difference inresult!1c. [10 points] What effect will raising the temperature have on this estimate of the flowstress? Hint: consider what happens to the shear modulus.The flow stress will tend to decrease with increasing temperature because the shearmodulus decreases with temperature (in almost all materials). Thermal activation ofdislocation motion (both conservative and non-conservative) will also lower the flowstress. Either of these is an acceptable answer.1d. [20 points] In a single crystal tensile test, the orientation is given as [259]//tensileaxis. Calculate the tensile yield stress based on the critical resolved shear stress that youobtained 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 todetermine the tensile yield stress because it determines which slip system will beactivated first. The direction [259] is approximately in the center of the standardstereographic triangle (i.e. 001-011-111).A suggestion for how to proceed is to use a spreadsheet (e.g. excel) and make alist of all possible combinations of slip plane (111, -111, 1-11, -1-11) and slipdirection (e.g. 111 is orthogonal to –110, -101 and 0-11), taking only positiveversions of each (unit) vector. This will give you a table with 12 rows, one foreach slip system (and 2 columns). Then calculate the dot products of the tensileaxis with each pair of plane+direction in turn in order to obtain cosf andcosl, respectively (2 more columns). Then you can calculate the Schmid factor ascosf*cosl (1 more column). Finally, identify the row with the largest absolutevalue of the Schmid factor in it (i.e. positive or negative). {You can expand thetable to include the negatives of each slip direction in addition: this will give you24 rows (e.g. 111 is orthogonal to –110, -101, 0-11, 1-10, 10-1 and 01-1). If yougo with the 24 row version, you will find that you obtain a pair of positive andnegative Schmid factors for each pair of positive and negative slip directions.This positive/negative pairing corresponds to positive and negative directions ofslip.}Answer: The largest Schmid factor is 0.49. If you used only 12 slip systems (nonegatives) then you would have had to take the largest absolute value of theSchmid factor. Negative Schmid factors simply mean slip in the oppositedirection on the same slip system. If you used 24 systems, then you can take thelargest 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 yieldstress of the polycrystal. You may find it helpful to draw a stereographic projection andnote the locations of the slip direction and plane on it that satisfy the maximum Schmidfactor criterion. Obviously the symmetry axes give an ambiguous result in that severalslip 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 shouldbe 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. Thereforethe polycrystal should yield