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 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 ij 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 zplane 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 1 3 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 a device e g by balancing all the vertical and horizontal components to make sure the device will not move In the case of stresses inside a material we resolve