This page contains questions and answers on a variety of topics, mostly from the problem sets. These discussions originally took place in the course forum between students, the teaching assistant, and the professor. Users with Javascript enabled can click on each question to see the answer or further discussion. Click again to hide each thread when you're done.

**Question 1:** Problem 1-1d

**Question 2:** Problem 1-3

**Question 3:** Problem 2-1, Young's Modulus and Poisson's Ratio

**Question 4:** Problem 2-2b

**Question 5:** Problem 3-1c

**Question 6:** Problem 3-2c

**Question 7:** Problem Set 3, Mohr's circle definition of alpha

**Question 8:** Problem 4-2a

**Question 9:** Problem 5-1b

**Question 10:** Problem 5-1c

**Question 11:** Problem 5-1e

**Question 12:** Problem 5-2a

**Question 13:** Problem 6-2

**Question 14:** Questions about toughening

**Question 15:** Prime stiffness and compliance tensor

**Problem 1-1d:**So I have designed a new kind of bike spoke, but no matter how hard I have looked and researched I am still confused about**local buckling**which I know shall occur if the walls of my spoke shall be too thin. The only equation that I have found that quantifies local buckling is F_{local}= k(π)Et^{2}where t is the wall thickness. However it was not made clear what k was, is this the stress concentration factor? I wanted to just check that my new spoke was not succeptible to local buckling.**TA:**Take a closer look at your recitation notes. That should help.**Student 2:**I think that what you are looking into is buckling of a cylindrical shell. Professor Wierzbicki teaches a class on plates and shells called 2.081. His notes have the specific buckling analysis of the problem that you are thinking of: Plates and Shells Notes (PDF)A good local buckling calculation is on page 84 of these notes. Also, "Plates and Shells" by Timoshenko is an absolute classic in this field. It will surely have the answer.

To specifically answer your question, the k is not really a stress concentration factor. If this k is the same one as I have seen in other sources, I suspect that it is more of a factor that accounts for both geometry, buckling mode, and a bunch of various coefficients that fall out from the math.

I paid close attention in recitation, but I don't remember plates or shells being mentioned at all. I guess your original question persists: do the instructors of the course really want for us to evaluate the local shell buckling, or only Euler buckling? I don't know.

**Professor:**Good discussion. Euler buckling for the appropriate cross-sectional geometry and end-support boundary conditions is fine. Of course, Euler buckling assumes the column is long and slender such that the first mode of buckling (bowing out to make a sort of "U" from the endpoints) is favored; you should check if this assumption holds for the length and diameter of your spokes.The term "k" you found can often be represented by different letters, but Student 2 is right that is dependent on the boundary conditions: are the two ends fixed, or is one fixed and the other free to displace laterally, etc. It represents the effective length between two points along the beam that have zero moment. If both ends are pinned, for example, this length is exactly the length of the beam (since ideal pins cannot sustain a moment).

**Student 2:**I think that Professor Van Vliet is still thinking of Euler buckling when she posted her reply whereas the formula that Student 1 posted is really a local shell buckling formula. Eq. 382 from the 2.081 notes says critical stress = 0.605*E*thickness/radius. Taking the area as the thin shell approximation Area=π*D*thickness, the formula can become F=2*.605*π*E*t^{2}, which is identical to Student 1's equation with the assumption that k=1.21. Professor Wierzbicki found the factor 0.605 by finding a minimum critical buckling load as a function of the buckling mode. The function that he minimized had geometry in it and accounted for simply supported boundary conditions. This is why I said in my initial post that the factor k falls out from "both geometry, buckling mode, and a bunch of various coefficients that fall out from the math". I forgot to mention boundary conditions on that short list, although the boundary conditions in this case can be indirectly accounted for in the buckling mode.Also, the constant in Euler's buckling equation accounts for both boundary conditions and buckling mode. For example, in the pinned-pinned condition, if you encounter the unlikely but not impossible scenario that the column buckles into the shape of a full sine wave rather than the typical half sine wave, then the pinned-pinned condition is still satisfied, but the leading constant is different.

**Professor:**Yes, you're right I was still thinking of and talking about Euler buckling there. It is a global buckling of the structure, even though it may appear to initiate locally. The criteria you are discussing, though, is not a large departure from that. Euler, strictly speaking, considered all modes of column buckling for a solid, slender column. The "critical buckling load" statement of Euler does consider the first mode, but higher modes (full sine, and so on) are possible if the modes that require lower critical loads are repressed. Your 2.081 notes consider the annulus cross-section, and the factor of 1.21 is due to the boundary conditions.

**Problem 1-3:**I'm a little confused about problem 3 of the homework. According to the table a high compressive load results in larger Radius (parallel to load) between the first peak and center of the x-ray pattern. Is this radius DIRECTLY related to interatomic spacing? If so, it seems that loading in compression actually lengthens the material. Is this correct?**TA:**The radius of the diffraction ring is directly related to atomic spacing by Bragg's Law, but they are not equivalent.**Professor:**Also, in a crystalline material, it is clear that tensile applied loads would lead to increased interatomic spacing in the tensile direction; in other directions, one would need to consider Poisson's effect. Recall that the structure of amorphous materials is, by definition, quite different from that of crystalline materials.

**Problem 2-1, Young's Modulus and Poisson's Ratio:**I am confused about what form of Young's Modulus and Poisson's ratio you want us to plug into the stiffness tensor. Is there a rigorous way to do this? Do you want us to just define different constants for different directions? Also, are we supposed to be taking into account the composite nature (Ch. 6 of Courtney) of the material, or just the symmetry/anisotropy?**TA:**The key to the problem is to treat the fibers and matrix as isotropic materials (maybe not an obvious assumption). Therefore there are values of Young's modulus, shear modulus, and Poisson's ratio for the fibers and the matrix. Using these values, you can describe the stiffness tensor of the composite by analyzing the stress and strain along different directions. Doing this requires an understanding of the symmetry of the composite.**Professor**: Remember that if a material is described with only 2 independent elastic constants (e.g., E and ν), it is implicitly isotropic even if this is not stated explicitly. One can idealize a material in several ways, ignoring any length scales and anisotropies (isotropic) or considering them to various degrees of accuracy (anisotropy described fully by the stiffness or compliance tensor).Composite materials can also be considered this way: looking at the material from a distance, it will appear isotropic (or at least have some isotropic bounds, depending on if it is deformed along or at some angle to oriented filler phases). Closer up, the anisotropic geometry of the materials clearly breaks the symmetry, even if those material phases are themselves considered to deform in an isotropic manner.

**Student 2:**It sounds like you're implying that the composite material in the problem is isotropic - but I assume that's just misinterpreted...with the fibers all aligned, it should be transversely isotropic, right?**Professor:**No, the composite material is not isotropic; the components of the composite material can each be treated (separately) as isotropic. At the composite level, it is definitely anisotropic. In fact, the conventional treatment of composites gives "bounds" for properties such as the stiffness measured normal to and parallel to the fiber long-axis. These bounds are given in terms of "E" which, since that's a mechanical property specific to isotropic continua, can create confusion.

**Problem 2-2b:**The question asks us to characterize the materials focusing on elastic module and relaxation time. I am confused about the relaxation time - I thought that this only corresponds to the time taken for material to return to 0 strain. However the data given has no data for unloading of the material...?**Student 2:**I think the idea is that every 0.5s, the user takes a step, and the average strain is recorded. By looking at the residual strain (and its change) after each step, we can determine the relaxation time for each material qualitatively. One material relaxes far faster than the other.**Professor:**Also, relaxation time and characteristic time are often used interchangeably, even when the experiment itself is not a "relaxation test" (removal of load and subsequent recovery by material). This is because:

(1) the quantity still represents the time required for the viscoelastic material to respond and adjust to the current stress state; and

(2) that characteristic time is fixed by the values of the springs and dashpots, and is the same value regardless of the applied stress/strain state.**Student 3:**I have discussed this with some of my classmates, and we are all confused as to what the test data in Problem 2 is supposed to represent. Above it is suggested that this is the response to some cyclic loading while some of us speculate that it is the response to a constant applied stress. Can Professor Van Vliet please clarify? Also, I can't figure out how to get elastic modulus information without knowing how stress changes with strain, but there is no stress information available in the problem statement or spreadsheet.**Student 3:**I have worked this problem through, and I am pretty sure that it is the strain vs. time response to a constant applied stress, and definitely NOT the response to a cyclic load of 2 Hz. However, in order to get the elastic modulus, I will need to know the constant applied stress. I can make estimates based on body weight and area of a foot, but this amount can vary widely between a small woman with big feet and a heavy man with small feet. Can we know the stress?**Professor:**How many small women do you know with big feet? Anyhow, you can assume the applied stress through justified assumptions (eg, as you suggest, the weight of a person and size of his/her foot). It's common in such cases to claim upper and lower bounds for these values. Also, I think you can figure out what the data represent by graphing it and thinking about how stress/strain change during walking at a typical pace (frequency).

**Problem 3-1c:**I am a bit confused about the intention of 1c. My understanding of the problem is that the loading will be something like F=0.5*F_{max}*(1-cos(ψ*t)) - i.e. a force control test that ranges from 0 to some F_{max}. Because the material is essentially "perfectly plastic", there is an ambiguity as to what it will do when it reaches either the maximum or minimum force. (This ambiguity can be seen for any case in which the actuator position is being controlled based on force feedback, but there is a very weak relationship between force and position). I think we can assume that the material strains plastically and stably when it reaches maximum force. Do we assume the same for the minimum force, or do we assume that the unloading phase is only elastic?**Student 1:**To elaborate... Consider a force-control test of an elastic-perfectly plastic material in tension. The material yields when the force on the specimen exceeds, say, 12.5 kN. You tell the machine to pull on the specimen until it reaches 12.51 kN, then return to zero, then repeat for 2 more cycles. However, when the material yields, the highest the force will go is 12.5 kN, no matter how long or fast the machine pulls on it. Therefore, the machine will become unstable, and it will pull on the specimen until it breaks, trying in vain to accomplish the impossible 12.51 kN. It will never do the other two cycles.On your second test, you foresee this happening, so you tell the machine to pull on the specimen until it reaches 12.49 kN. Then the machine loads the material for three cycles in the elastic region, and it is very boring test with no plasticity or fracture.

On the third test, you know EXACTLY what the perfectly plastic yield point is (12.50 kN), and you tell the machine to cycle from 0 to 12.50 kN three times. I really don't know what the machine would do at this point. It could go up to 12.50 and strain the material until fracture on the first cycle, or it could stay elastic the whole time.

My understanding of problem 1c is that we are being asked to explain what will happen in exactly this situation. Based on the fact that the problem says "the minimum load being 0 N", I assume that the problem is implying that this is a force-control experiment, and that it is in tension. If the problem were in compression, then it would be solvable because the force in the plastic regime increases monotonically due to area increase. However, in tension, the cross-sectional area decreases, so the force is elastic-perfectly plastic at best, and softening at worst. I don't think that Professor Van Vliet wants us to solve difficult problems with respect to machine control systems, so I think that I must be mis-understanding the problem. I was wondering if Professor Van Vliet could clarify the problem. E.g. Is this cyclic loading to fracture in compression? Is this a position control test? If this is a force control test in tension, then what can we assume that the control system will do when it reaches the maximum and minimum loads?

**Student 2:**I was always of the opinion that we have to consider compression where the increase in cross-sectional area would account for the increase in force, because the simplest solution could be the right solution. I am also curious about what you mean by position control test? Loading up to a certain strain? How is that different from force control? Aren't you controlling the load in any case?**Student 1:**In a position control test, you tell the machine to extend at a certain rate, e.g. 5 mm/min. In a force control test, you tell the machine to extend at a certain force rate, e.g. .5 kN/min. The position control test requires either open-loop control or relatively simple control systems that say "if I am going too slow, then I speed up; if I am going too fast, then I slow down." The force control test typically requires a somewhat more sophisticated control system whose parameters depend at least in part on the stiffness of the specimen and machine. In this case, the control algorithm says "if the force is too low, then I extend the position; if the force is too high, then I decrease the position". Knowing how much to extend or decrease the position depends heavily on the force-displacement characteristics of the specimen, and machines sometimes need to be "re-tuned" for different materials or specimen geometries. Clearly, if force doesn't change with position (as in an elastic-perfectly plastic tension specimen that has yielded), then the logic of "if the force is too low, then I extend the position" breaks down, and the machine becomes unstable.Thanks for the comment that you think the problem is in compression. I think that this is the easiest problem to do, and I will assume that this is the correct approach... at least until I see something posted otherwise.

**TA:**I apologize if the question misled you. You are exactly right about how an actual test with a machine would go given the elastic-perfectly plastic character of the material. However, the intent of the question was simply to make you think about how the stress-strain curve would look if you loaded to a stress above the yield stress and then released the load. If you understand this, then you know what will happen if you reload the specimen to a stress above the yield stress. Perhaps this material is not the best example because it does not strain harden which (as you pointed out) implies that the stability of the plasticity is difficult to control experimentally.**Professor:**This was a good discussion, and compression is the easiest case to consider when debating the effects of position vs. load control. I asked the TA to take care of discussing this point with you because it was tangential and I was working on answering other questions. Now I have time to chime in.Position control and load control are very different in terms of the mechanical instabilities accessible in a material. In experiments of uniaxial loading, load control (feedback loop based on the load registered by the strain gauges in the load cell) is typically used; this is because one is trying to often control the loading rate and to accurately determine the stress to yield or otherwise fail the material.

In simulations based on finite element modeling, however, displacement control (feedback loop based on the displacement registered by a strain gauge, optical encoder, LVDT or other devices placed within the load train) is typically used. This is because such simulations use the algorithm we have repeated in class many times (almost a mantra): displacement incurs strain; strain incurs stress defined by the constitutive laws appropriate to the material.

For a material that is perfectly plastic, the onset of yielding is a severe instability: any further strain incurs no further stress, so there is no unique solution to the stress corresponding to a given strain (from a simulation point of view); and reaching the strain to failure can occur very quickly because that strain is attained with no increase in applied load (from the experiment point of view).

A great example of this is in dislocation nucleation as measured via contact loading of a surface. Nanoindenters operate in load control, by inherent design of the instrumentation, and register dislocation nucleation and motion as a "burst" of displacement with no concurrent increase in load. Simulations of this experiment operate in displacement control (tell the program to move the indenter toward the surface at a constant velocity), and register dislocation nucleation and motion as a sharp decrease in load with no concurrent increase in penetration depth.

**Problem 3-2c:**In Problem 2c we're asked to plot two series of stress-strain data onto yield surfaces, what are the units of stress in the Excel file?**Professor:**The TA can answer this directly, as he generated those data and we should have stated units in either the .XLS or the problem set document. To figure this out for yourself, though, you would consider the fact that yield stresses of engineering materials certainly vary but are typically less than 1 GPa and approximately 100s MPa for alloyed or highly defected metals. Having these order of magnitude values in mind, just as in the case of Young's elastic moduli for metals, ceramics, and polymers, is helpful in terms of spotchecking your and others' work.

**Problem Set 3, Mohr's circle definition of α**The PS3 statement of how to construct the Mohr's circle to consider the Mohr-Coulomb yield criterion was designed to make you all create the same basic Mohr's circle, but has caused some confusion. It is important to note that "α", the friction angle, is a property of the material; the material is also defined by a value of cohesion. There is one value of α for a given material, just as there is one E and one G for a linear elastic, isotropic material.α is defined by the tangent to a Mohr's circle at the point on the Mohr's circle called (σ

_{y}, τ_{y}), taken at the point where that tangent crosses the τ = 0 (horizontal) axis of the Mohr's circle. This tangent does not need to cross at zero, but in fact crosses at some constant number related to k. It is the angle defined by that tangent and the horizontal.(Physically, as stated in the problem, it is the angle that a granular material would naturally acquire if you placed a cylinder of that material on a horizontal plane (grains slide down to make a little pile according to the amount of friction between the grains; hence, it's called the friction angle). The Mohr's circle can be thought of as a stress state required to induce that angle.)

**Student 1:**I'm pretty sure that the Drucker and Prager equation you gave us is incorrect... Instead of the 2nd term being (α x I), it should be (γ x I), where γ is a function of α (other sources use α in the equation, but use another symbol (i.e. γ or ω) for the friction angle): source. However, even with this reference book, I can't get this relationship to get an appropriate plot.**Professor:**The definition of Drucker-Prager is correct as stated in PS3. However, two caveats:(1) This was not Drucker and Prager's original definition. I restated it for you so that "α" and "k" take on the same physical meanings as they do in the Mohr-Coulomb criterion: α is the friction angle and k is the cohesion shear stress or cohesion strength. This is the most common way to state this criterion. You will also find it stated with a different constant which they called "a". This is because their criterion requires measurement of theoretical "k", which is hard to do experimentally; "a" accounts for the difference between the compressive and tensile yield stresses.

(2) Drucker-Prager is also stated as

f = √(J

_{2}) + (α)(I_{1}) - kwhich is the same as stated on your pset except now there is a "+" in front of (α)(I

_{1}) instead of a "-". The reason for this easy to understand, and comes up often in mechanics: sign conventions. D-P wants to reflect the fact that the material yields differently in tension and compression. The stress tensor invariant I_{1}would be a negative number if one puts the minus sign next to the diagonal elements of the stress tensor to reflect compressive stress; it would be a positive number if one neglects the minus sign or considers compressive (negative) hydrostatic stress to be stated as positive pressure.Note that neither equation is right nor wrong, and both statements of D-P yielding are correct. The important part is to make sure the criterion and its evaluation make physical sense: they should reflect the fact that yielding is altered by the magnitude of hydrostatic stress and occurs at lower magnitudes of stress in tension than in compression for a D-P material.

**Problem 4-2a:**In this question, is "r" the atomic radius of solute? I do not think it makes sense. I think r is the atomic radius of the matrix atom, then r*(1+δ) will be the radius of the solute. Refer to Courtney's book, page 189 about solid solution strengthening.**Student 2:**I think δ should be replaced by e in order to be dimensionally correct. The δ typically has units of length, so r(1+δ) would have units of length squared. In the text, the solute atom has radius of r(1+e). And anyway, the question asks we express the volume change in terms of the induced strain e.**Professor:**That is correct. δ is necessarily unitless, a percent mismatch of radii, and is thus equivalent to strain.

**Problem 5-1b:**I know Coble creep is more grain-size-dependent than NH because of the higher power of d in the constitutive equation. But how do you discern the difference between the two in looking at strain rate v. stress data?**Professor:**There is another difference between them, which is the mechanism/corresponding activation energy. These are often tabulated for specific materials in papers, textbooks, and online databases.

**Problem 5-1c:**You want a schematic here for the varying grain size, right? ie., how does strain rate v. stress behavior generally change for increasing grain size?**Professor:**Good question. There will be a strong effect of grain size on the strain rate due to creep mechanism, but there could also be an effect of grain size on strain rate even with a given creep regime. To consider whether this effect is observable compared to the creep mechanism effect, you'd ask yourself whether others have reported strong effects of grain size on stress-strain responses and/or yield strengths and strain hardening coefficients for this class of material (crystalline oxide) over this range of grain size (1 micrometer and larger). The answer is generally no, they do not, but you should look into it and cite your findings.

**Problem 5-1e:**I guess there is no information about G and T_{m}of the polycrystalline oxide in the problem set. Is it O.K just to draw the map based on σ and T only?**Professor:**Remember that these graphs are often rendered with dimensionless variables, i.e., σ/G and T/T_{m}.

**Problem 5-2a:**Do we use the stress transformation equation for maximum in-plane shear stress, given σ_{x}=10 MPa, find that max shear stress is 5 MPa, and at T=-50C determine the corresponding dominant creep mechanism (i.e. power law)?**Professor:**Yes, anytime you're given a particular stress state and asked about maximum conditions (shear stress, normal stress, planes on which those occur), you rely on the stress transformation concepts. Once you have stresses of the type stated on deformation mechanism maps, you can locate your (temp, stress) position on those maps.

**Problem 6-2:**I think the y axis represents K_{IC}not K_{I}, because it's the value of K_{I}when fracture occurs. In lecture, we learned K_{IC}is a material constant, so why in this problem does K_{IC}depend on geometry?**Professor:**Correct, the axis is labeled "fracture toughness" by the authors, and we defined it for you in the problem statement as the stress intensity factor when the sample breaks. This is different from K_{IC}, which is the plane strain critical stress intensity factor under Mode I loading under the conditions we defined. You are correct that K_{IC}is a property of the material.

**Questions about toughening:**First question is about lecture 18. You showed us brittle fracture pictures, and you said that grain size reduction makes the materials tougher in both intergranular fracture and transgranular fracture. Did I hear correctly? Then, I wonder why reduction of grain size makes toughening of materials in both cases.Secondly, in fatigue toughening I wonder how crack bridging works. I have already read Courtney about crack bridging but I can't understand it.

**Professor:**1. You heard correctly. Remember that toughening is always an increase in toughness as compared to some other specific case. You have to ask yourself, tougher than what?If the material fails by intergranular fracture, the grain boundaries are the least fracture-resistant part of the material microstructure. If the grains are large (d = 10 micron) the grain boundary surfaces will also be large (micron-scale paths), and a crack can propagate along that path without interruption. If the grains are smaller, the crack will have to change directions every time it hits a new grain boundary. If there is not enough stress in, say, Mode I loading to drive the crack to propagate along the new direction, it will stop propagating. That means it is tougher than a material of larger grain size (and same grain boundary fracture resistance).

If the material fails by transgranular fracture, the matrix is the least fracture-resistant component. The grain boundaries then serve as obstacles to cracks propagating from one grain to another.

**Professor:**2. Crack bridging: There are two types we discussed.(1) If there is a main crack that is propagating through the material, it may stop propagating when it encounters an obstacle (or when it blunts due to plasticity, etc.). If the applied load is maintained, the stress that builds around that immobile crack tip can be sufficient to propagate a new crack ahead of the main one. The uncracked material between these cracks holds the part together, toughening it compared to a material that had one big connected crack, but eventually these cracks may link up to form a big crack that leads to failure.

(2) Fibers / filaments within the material may be aligned normal to the crack faces and hold the crack faces together until the fracture stress of those fibers is attained.

**Prime stiffness and compliance tensor:**Someone in recitation today asked about if you would need to use a "prime compliance tensor" (i.e. transformed) when comparing a "prime stress" to a "prime strain". The answer depends on whether the compliance tensor is in reference to your prime coordinate system. In our example, the prime coordinate system was along the {100} family of planes. If I gave you the compliance tensor with respect to this coordinate system, no transformation would be required (i.e. no prime necessary). However, if the compliance tensor was that given with respect to the {110} family of planes (our original or old coordinate system}, then a transformation of the compliance tensor would be required.**Professor:**I just want to clarify why we discussed the "primed" version of the compliance tensor in class. (We wrote S'_{1111}in terms of unprimed S-tensor). The utility of the S- or C-tensor transformation is that it is an alternative means of identifying nonzero elastic constants based on material symmetry. As discussed in class, if we knew that a 90 degree rotation about the x_{3}-axis should produce no change in the measured stress inside the material due to an applied strain, we could use that fact to identify which S' components are exactly zero (because, if they were not zero, it would be impossible that the stress before and after the 90 deg rotation of reference axes would be of identical magnitudes).