My original intent was to give Rob a cautionary “heads-up” on a situation that he may not have been particularly aware of or have thoroughly investigated, and that could possibly have significant consequences. To illustrate how quickly the stress in the Heim joint shank can rise when it is subjected to bending loads, the FEA model was done with roughly his geometry and a single, plausible, load case. It seems to have achieved that end.
Since the focus was on the shanks, the details of the rest of the structure were of secondary importance, although the results do give an indication of the stresses in the other members. Without writing a treatise on FEA or structural analysis, the following may help in understanding what was done and what these results indicate.
Using static force analysis and the theory of elasticity, classical engineering equations allow one to calculate the stresses and deflections of simple structures by hand. For instance: if one had a cantilever beam projecting from a wall with a weight hanging from the end, the nature of the loads at the wall end and the resulting stresses can be easily calculated. As is intuitive, at the top of the beam there would be tensile stress and at the bottom there would be compressive stress. If one grabbed the beam at the end and pulled away from the wall, the resulting stress would be uniform tensile stress across the beam cross-section. If the end of the beam was twisted about its axis, the stresses at the wall would be more complicated, but can be calculated. Each of these loads will create a stress component at any location in the cross-section and if all three loads were applied the net result would be the (vector) sum of their contributions at each location across the cross-section of the beam.
In the same way, the stress components at the mid-point of the beam could be calculated, or at any intermediate point along the length. The beam could be thought of as a series of short beams all connected end-to-end. If the calculations were done, the distribution of stress all along the length and across the the section of the beam would be known.
If we were to analyze the beam using FEA, we would do essentially the same thing. An FEA model is composed of “elements”, which “model” a section of beam, and in this case assume the beam is of uniform cross-section. The endpoints of the beam section are defined by “nodes” located in space. If we are not all that interested in the stress results along the length of the beam the model could consist of a node at the wall, another at the other end, and a single “element” in between. This model would use the appropriate equations to calculate the stresses in the same manner as would have been done manually above.
If we were interested in how the stresses varied along the length of the beam, we would have to create a model that consisted of a whole series of nodes with an element between each. The calculated results would then give the stresses at each end of each element along the length of the “beam”. This requires solving a lot of simultaneous equations which, fortunately, computers are reasonably adept at.
In general, the stress state at one end of an element may be different from that at the other end. In the simple single-element beam above with the weight and axial tension load, the stress at the outboard end would be the nominal tensile stress from the tensile load, but the end at the wall would be a combination of the tensile load stress as well as the bending stress caused by the outboard weight.
An FEA program calculates and stores the various components of stress at each location, generally tied to the direction in which they act within the element. This can often result in a lot of information that is difficult to interpret or relate to the ability of the material used to withstand the load. Fortunately, for steel and many other ductile materials, the stress components can be mathematically combined into a single net stress that can be compared directly to the strength of the material. One of these is the von Mises (or various other names) stress. If desired, FEA programs also will make this combination calculation and present that stress as a result for the element.
So, to bring this together as regards the suspension model, in the interests of simplicity, it was made of single element members, since the details of what was going on within the members was not of particular interest--just the magnitude of the greatest stresses of the particular member. Thus, the maximum von Mises stresses were tabulated for each element. For the Heim joint shanks, the maximum can intuitively be assigned to the end where the bending load is maximum, at its juncture with the link to which it is attached. (This is confirmed by looking at the more obscure results of the analysis.) To summarize, the tabulated results are an effective net stress that can be compared to the material strength for that member, given the arbitrary load situation analyzed, and in the absence of geometrical stress concentrations.
Stress concentrations ---
As described above, the elements used in the analysis assume uniform cross sections. If the size, shape, or configuration of the adjoining member is not a continuation of the one in question, a degree of stress concentration will occur due to a mis-match of the stiffnesses of the two elements. The model is, in effect, an idealization consisting of “perfect” connections between the members. At the welded joints of the tubulars, clearly the member cross-section is disrupted and stresses will be re-distributed depending on the nature of the loads and material configuration at that location. To answer Rex’s question, no, the tabulated stresses do not take into account the variations that will occur at the welded joints of the X frame. Those were not the goal of the exercise, although considerably more elaborate modelling of the junctions would give accurate indication of what is going on there. However, the given results for the members in conjunction with various stress concentration factors historically developed for similar joints would give a fair indication of what could be expected.
For the Heim joint shank, a smooth 5/8” diameter cylinder was assumed as an approximation of the root diameter of an assumed 3/4” OD thread, but due to the sharp root of the thread that reduction may not fully compensate.
The attached two plots may partially address the stress distribution question that Rex was sort of fishing for. The VM0 plot shows the stress magnitude on the top of the members, and so is largely driven by the bending loads in the vertical direction. As would be expected, the vertical bending induced in the trailing elements of the X by the “rolling” moment from the offset (lower) lateral load applied to the “axle” predominate. (That is, the X is being twisted by the axle.) This plot shows the stress magnitude by the offset from the normal element line, and the color code. It is basically the result at each end of the element, with the middle portion merely being interpolated from either end. Also notable is that the stress at the top of the element in the right side Heim is fairly nominal.
The second plot, VM90, shows the stresses in the horizontal plane, on the side of the elements, which are largely the result of the lateral load itself. As can be seen, the stress due to this is rather low in the X since its members are acting substantially by carrying longitudinal, not bending loads. Also note that the front right Heim is showing high stress at its base due to its carrying substantial bending loads.
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As regards Interested Bystander’s desire to more simply categorize the potential severity of the consequences--that is pretty much up to who is designing/building/driving the thing. And again, the stress situation discussed is for an approximation of a contingency situation. What one is willing to accept may be different if you are going to the moon or to Floating Mountain.
Rex’s discussion of all-pinned-joints sounds similar to the “N” braced version mentioned a number of posts back. Although, it would have to be an N, not an X. Any roll-axis motion would induce bending in the X (as in VM0 above.)
And if IO was going to do it, as alluded to earlier, he would opt to separate the functions for tunability and sanity, and eliminate bending wherever possible.
