## General Mechanical

Topics relate to Mechanical Enterprise, Motion, Additive Print and more

#### Post Processing: How to determine the mode of failure for isotropic materials?

• Rameez_ul_Haq
Subscriber
Hello. Can anyone please guide me on how to determine the mode of failure for an isotropic material. This question is about post processing, which I am currently struggling with. For example, I have a steel, with some loadings and boundary conditions, and it is observed that some part of the steel has Von-Mises or Tresca stresses greater than the yield strength of it. Now I want to modify the geometry such that it wont fail. How can I do this? For example, if I can find if the failure is happening due to high bending stresses, or tensile/compressive stresses, or shear stresses, or buckling stresses, then that can help to have an idea how the geometry should be altered. n
• peteroznewman
Subscriber
nSteel is a ductile material, so the relevant stress metric to determine if the part stress has exceed the yield strength is the von-Mises stress, called Equivalent (von-Mises) in Mechanical.nThe corrective action in the part design depends on the geometry of the structure and the location and type of applied load.nFor example, in a cantilever beam type of problem where the stress is too high, you add material if the load is a force, but you remove material if the load is a displacement.nWhen the load is a force and the stress is too high, that means the structure is too flexible and by adding material, you can stiffen the structure which will reduce the deformation and reduce the stress.nBut when the load is a displacement and the stress is too high, that means the structure is too stiff and by removing material, you can make the structure more flexible, spreading the required deformation over a larger volume of the structure and reduce the peak stress.nA structure can have many requirements and if any one requirement fails, the structure has failed to meet its design requirements. For example, the structure might have a deflection limit. All stresses in the part can be below yield, but if one face has exceeded a deflection limit requirement, then the structure is too flexible. The corrective action for exceeding a deflection limit is different than the corrective action for exceeding a stress limit.n
• Rameez_ul_Haq
Subscriber
thank you for answering.nAre there any research papers or books which you can kindly recommend which explains how and where to modify the structure according to the loading and failure type. How to incorporate the effect of load paths and alter the geometry accordingly so that the stresses can be gently distributed without causing any failure?nMy question was actually this; for example I use Von-Mises stress, now that will be a combination of all kinds of stresses, but I want to know which type of stress (like tnesile or compressive or shear) at the location where Von-Mises stress is higher than the yield strength, has the most effect on the Von-Mises stress. Is there any option to know that in ANSYS?n
• peteroznewman
Subscriber
nYou could make six contour plots and look at the six components of stress that make up the von-Mises stress. The six components are directional, so you get a different value depending on which coordinate system you plot the stress in. It is generally more useful to transform the angle at which you are plotting the stress into one that causes the shear components to be zero. That is what the Max Principal and Min Principal stresses are. They are the stresses that result from choosing a coordinate system where the shear stress is zero. It is also helpful to plot these as a vector graph so you can see the angle of the coordinate system used to compute the principal stress, since the angle changes at every point in the part.nThe simplest advice on how to think about a structure to reduce stress caused by applied force is this: Axial Good, Bending Bad. That means design parts so that the load path travels axially through the part. Avoid part designs where the load tries to bend the part. nThe simplest advice on how to design structures that have a required bending displacement, such as a cantilever spring, is to use taper. Either taper the thickness of the part, or taper the width of the part so that the fixed support has more material than the tip where the displacement is enforced. Make the part longer to reduce stress. Of course, there is usually a competing metric such as reaction force, so that becomes an optimization problem to minimize stress subject to a required force at a required displacement (or spring rate).nI don't know of a book or paper that covers this kind of knowledge.n
• Rameez_ul_Haq
Subscriber
''It is generally more useful to transform the angle at which you are plotting the stress into one that causes the shear components to be zero'', why ? Because as far as I know, ductile materials fail in the direction where the shear stress is maximum. Why do we need to check the max and min principal stresses and their directions for a ductile material?n''That means design parts so that the load path travels axially through the part. Avoid part designs where the load tries to bend the part'', Is this because the magnitudes for max and min principal stresses are going to be greater in the bending case as compare to a axial loading case, since there will also exist an initial shear stress on an element within the structure in the former, while no initial shear stress on the element in the latter.nYou mentioned this, ''Make the part longer to reduce stress'', how will it reduce the stress Sir? Does STIFFNESS and RIGIDITY of a structure has an effect on the stresses observed within the structure? (because we all know stiffness and rigidity has an effect on strains definitely). Because increasing the length of beam will decrease its stiffness which means the displacements would increase, but how would it decrease the stresses experienced by the cantilever beam. Is it because the moment arm has increased (which, by the way, should cause the stresses to soar up), or does it actually have to do with the change in stiffness of the structure?n
• peteroznewman
Subscriber
• Rameez_ul_Haq
Subscriber
,so basically what you said is that there is no difference (or are very close to each other) in the directions where the actual load will travel through the part and the directions of the Maximum principal stress within the part? And why should we ignore the directions of minimum principal and only be concerned about maximum principal? Or we should take into account the directions of that principal stress whose magnitude in greater than other?nAnd also, speaking my mind out right now, I actually never understood that how does the trusses in a bridge, or somewhere else, understands that they are supposed to carry only the axial forces. For instance, have a glance on the picture below.nconsider JOINT A. There are two reaction forces i.e. Ha and Ra are acting on joint A. Can't the force Ha or Ra be taken completely by the inclined member so that it bends as well Why do we assume that the members only take forces along the axial directions?
• peteroznewman
Subscriber
why should we ignore the directions of minimum principal and only be concerned about maximum principal?nMost materials fail in Tension at a much lower level of stress than Compression. The highest level of tension is shown by Max Principal stress. The highest level of compression is shown by the Min Principal stress. The directions of these two stresses are orthogonal.nA true truss structure is assembled using pins at the ends of each strut. The pins prevent any bending loads from entering the strut. Only axial forces can enter the strut. Some bridges have struts welded to plates at the joints so that bending loads could enter, but then it is no longer a true truss structure.nIn a true truss structure, you can compute by hand the tension and compression in each strut from the applied loads and the reaction forces at the supports. In the figure you show, the applied load P is vertical, so Ha = 0. Once you weld the joints, it is much more difficult to compute the stress in each strut by hand because bending will be allowed, so a FE model would be used to compute the solution instead.nWhile exceeding the strength of the material is a failure mode for struts in tension, there is an additional failure mode for long slender struts in compression: buckling. ANSYS can do a linear Eigenvalue Buckling analysis to compute the critical buckling load.n
• Rameez_ul_Haq
Subscriber
thank you for your clarification.n[so basically what you said is that there is no difference (or are very close to each other) in the directions where the actual load will travel through the part and the directions of the Maximum principal stress within the part?]. Can you kindly give your affirmation on this statement as well please?n
• Rameez_ul_Haq
Subscriber