3.032 Class Questions Re: Quiz 2 Topics Fall 2007 Note: These are responses to class members’ questions in order of emails received, not necessarily importance. My responses are intended to supplement your class lecture notes and reading material. 1. Please go through the shrimp at the base of the ocean problem again. I am not clear the difference between stress and shear when it comes to the death of the shrimp. That was a great example you brought up. A material or an animal within the ocean is subjected to pure compressive hydrostatic stress (normal stresses in x, y, and z are equal in magnitude). The maximum normal stress criterion for plastic deformation predicts that a material will yield if the normal stress reaches the value obtained when the material fails under uniaxial tension/compression. In other words, the material will fail if ANY of the normal stresses reaches yield strength sy. The uniaxial compressive stress required to yield or kill a shrimp (0.1 MPa), so this criterion predicts that a shrimp would yield/die if it reached a depth in the ocean where the pressure reached 0.1 MPa. However, shrimp live happily at depths where pressure is 10 MPa, so this criterion does not accurately capture the “yield criterion” that predicts plastic failure of shrimp. We discussed that the reason for this failure is that many materials including the biopolymers that make up a shrimp fail not due to high normal stresses but to high shear stresses. From Mohr’s circle, it is easy to see that hydrostatic stress fails to generate shear stress on any plane inside the material: this is a principal stress state where all 3 normal stresses are located at the same point on the σ-axis, so you cannot draw a circle to define planes on which the shear stress is nonzero. 2. Principle [sic] stress, principle [sic] shear: What does this mean in an engineering system? If I have a ceramic I know has a yield stress of x MPa, how does a 2 x 2 matrix of stress and shear indicate how and in what orientation I can implement it? Engineers need to be able to determine the principal state of stress or of strain in a material to identify the maximum/minimum values that normal/shear stresses/strains will attain inside the material. As we learn from the yield criteria, if these maximum values reach the magnitudes required to yield the material, the material will deform permanently (plastically) rather than reversibly (elastically). This often leads to functional failure of the structure or device, and computation of the principal stresses corresponding to a triaxial stress state is the fastest way to determine whether yielding will occur according to these criteria. By “2 x 2 matrix of stress and shear”, I think you mean a matrix σij that includes both normal and shear stresses; if it is a 2x2 matrix then you are assuming plane stress: the only nonzero normal/shear stresses are contained in one plane (say, x-y plane), and there are no normal/shear stresses in the orthogonal plane of the material (say, the z-plane). This is a reasonable approximation for thin sheets of material, where z is the through-thickness direction. If you had a ceramic of known yield stress x MPa and knew your application would place this ceramic under this plane stress state, you would want to immediately determine the principal stress state. Why? First, you could then determine whether these stresses are sufficient for yielding to occur, according to your choice of yield criteria. Second, this would tell you the orientation and magnitude of the maximum normal stress. Although we’ve not yet covered fracture in 3.032 yet, ceramics are typically brittle materials that fracture rather than plastically deform, and the maximum normal stress criterion does a good job at predicting the fracture stress of such materials. If the maximum normal stress in the ceramic reaches the fracture stress of your ceramic, you know right away that the ceramic device will fail and you would choose another ceramic of larger fracture stress. If the maximum normal stress is below the fracture stress of your ceramic, you’d still want to consider modifying the ceramic via processing to either increase its fracture stress or adding fibers that would be aligned to block growth of cracks. Thinking back to the pressure vessels/hot dogs, you can intuit that the cracks would run perpendicular to the maximum normal stress, and you would then align these fibers parallel to the maximum normal stress orientation (the principal stress axis orientation) to be sure you’d block such cracks. 13. Do components like fibers on a silicon substrate really undergo stress tensors like in that problem set? How? Do engineers at intel and other device companies really look at stress states like that? Yes, engineers at electronic device companies like Intel in fact consider such stress states on silicon. Typically, such engineers are concerned with triaxial thermal stresses when these devices reversibly heat and cool. If the silicon substrate is both heated and loaded mechanically, which is common in such devices, both normal and shear stresses are generated within the Si wafer. Interconnect lines of metal and device insulators of oxides are so thin on such devices that, at the interface where those tiny lines of materials join to the silicon substrate, the strains generated in those small-volume “fibers” or films are exactly equal to the strains in the Si. 4. Why do I want to rotate a stress state x degrees? What does this physically mean in an engineering system? We’ve discussed this through several different examples in class and recitation, as well as in Lab 1. We are not rotating a stress state, but rather resolving the stresses or transforming the stresses onto a new set of coordinate axes. Coordinate axes are arbitrary choices to analyze a real engineering system; we often pick axes that align with the shape of the object or the orientation of the applied stresses/strains, but that is purely arbitrary. If we were to draw a different set of coordinate axes rotated by some angle theta, we’d come up with a different combination of stresses/strains. This is the same exercise as resolving a force vector that is at some arbitrary angle into its “vertical” and “horizontal” components. We have not rotated the force, just expressed that force in terms of a new set of axes. This makes it easier for engineers to analyze all the complex forces acting on