

September 18, 2020 at 4:14 pmRameez_ul_HaqSubscriberThere is an option of 'History Data' under the fatigue tool, Loading option. Can anyone give me a further insight into which type of data should be entered into here? I have a cyclic loading condition where the amplitude of the loads, its frequency and the maximum and minimum loadings are changing over time. However, I have conducted a static analysis for a approximately average load. But I want to know the fatigue behavior under the conditions I have mentioned.nCan anyone enlighten me on how can I perform this?n

September 18, 2020 at 6:26 pmDaniel ShawAnsys EmployeeHistory data is an ASCII file that contains load multipliers. For example, if you specify fully reversed Constant Loading (rratio = 1), the load (actually the FE calculated stress) varies from 1x the calculated value to 1x the calculated value. You could achieve the same result in the Fatigue Tool (FT) using single column history data file that had a 1 in the first row and a +1 in the second row (order is not important). However, with history data, you can have more than 2 multipliers. For example, if your history data file was a single column with these values: 1,1,2,3,0,2,4. The stress history would be 1x, 1x, 2x,3x, 0x, 2x, 4x the FE calculated stress. The FT calculates the fatigue damage produced by the entire loading history.n

September 18, 2020 at 6:52 pmpeteroznewmanSubscriberThe history data entry in the Fatigue Tool is how you input the loadtime data. For example if a force is applied at the end of a cantilever and it cycles many times between a minimum and maximum value, the Fatigue Tool will perform rainflow counting and bin the data into amplitude bins with the cycle count for each bin.nYou create a Static Structural model and apply a unit force at the end of the cantilever. The fatigue tool will multiply the stress in the unit load case by the amplitude of each amplitude bin to get the different levels of stress and compute the damage by applying the number of cycles in each bin using Miner's Rule.nFor example, if your loadtime history data for the end of the cantilever cycles between 63 and 71 N of applied force, you make a Static Structural model and apply 1 N to the end of the cantilever. This is a linear analysis, so the stress at 71 N is just 71 times larger than the stress at 1 N.n

September 21, 2020 at 8:47 amRameez_ul_HaqSubscriberReplying to Danielshaw and Peter,nIf I also want to take into account the time taken for each loading to occur, is it possible to add this parameter to the ASCII file which is supposed to be attached to the History data? For example, 1 occurs at 0 sec, 1 at 0.5 sec, 2 at 1.5 sec, then 3 at 2.25 sec and so on. Or frequency of the load blocks written in the ASCII file doesn't affect the results in ANSYS?nPlus, it would be extremely helpful if you can attach an example ASCII file which shows the load factors as you have mentioned, for example 1,1,2,3,0,2,4.nAnd one last thing, if I have several forces and moments applied to a part, but I want to see the fatigue response of the whole part subjected to only few of them being cyclic loads, how can I perform this? Say I have forces of 10, 15 and 20 N and moments of 50, 60 and 70 N.m, I only want to make 15 N and 70 N.m being cyclic, but at the same time I want the part to remain under the infulence of other constant loads as well, how can I perform fatigue analysis for this condition?n

September 21, 2020 at 2:37 pmRameez_ul_HaqSubscriberAnd also, is it possible to change the boundary conditions for each of the load factor? Like for a load factor is 1 the boundary conditions is fixed support, but for the next load factor of 1.5 the boundary condition becomes a displacement entry, for the same scoped geometry. Is there an option like this either in ASCII file or somewhere else?n

September 22, 2020 at 6:35 amRameez_ul_HaqSubscriberCan someone answer these please?n

September 22, 2020 at 9:00 pmpeteroznewmanSubscribernWhy do you care about the time taken for each loading? This is a linear statics analysis. Time means nothing here. There are no dynamic effects. There is no creep in the model. All that matters are the ups and downs in the load history.nHere is the load history for 1,1,2,3,0,2,4. Copy the lines below and paste them into a text file.n1n1n2n3n0n2n4nFor the case of two alternating loads in the presence of other static loads, duplicate the model and run just the static loads in one model, and just the alternating loads in the other model, which can have the fatigue tool included. Find the location in the alternating loads model with the maximum value of Equivalent stress. If the alternating loads are fully reversed, then the mean stress is zero in that model. In the static loads model, obtain the Equivalent stress at the same location where the maximum stress was found in the alternating model. Plot a point on a graph where the X coordinate is the stress from the static loads model and the Y coordinate is the stress from the alternating loads model. Label the X axis Mean Stress and the Y axis Alternating Stress. Draw a Mean Stress Failure Theory line to create the Safe/Fail boundary. The Goodman line and the Gerber line are typically used. Most graphs call the Y axis Alternating Stress, but in the figure I found below, they used Variable stress, but it is the same thing.nIf your point is below the line, your part should have infinite life.n

September 29, 2020 at 5:19 pmRameez_ul_HaqSubscriber''Find the location in the alternating loads model with the maximum value of Equivalent stress'', you mean to say that find the maximum value of equivalent ALTERNATING stress right? And what if the alternating load case have a mean, how can I include its effect? Should I add it to the equivalent stress gotten from the static load condition? nAnd one more question, which I think is critical, for example I apply a load of 200 N to a structure, and I make the loading ratio as 0.5, meaning the stress is oscillating between 200 N and 100 N, then if I add 'equivalent alternating stress' under the fatigue tool, and say I get a maximum value of 500 MPa at a specific location within the structure, does this mean that this 500 MPa is alternating between 500 MPa and 500 MPa? Since I have chosen a loading ratio of 0.5 instead of 1, I don't think it should mean the alternating stress is oscillating between 500 MPa and 500 MPa. What are your thoughts on this?n

September 29, 2020 at 5:42 pmRameez_ul_HaqSubscriberAlso, consider my structure Aluminum Alloy. If I use a loading ratio of 0.5, and go for the Soderberg criterion, the ANSYS will take the resulting number of cycles from SN Curve for which R ratio value, like for 1, 0.5, 0, or 0.5 (as seen from the engineering data)? How will it know which SN Curve it has to refer to? Or it will just directly consider the SN Curve for R ratio value of 1? If yes, then why?n

September 29, 2020 at 6:13 pmpeteroznewmanSubscriberThe stress in the alternating loads model is an alternating stress, of course. You divided the loads between two separate models. You use Design Assessment to create a Solution Combination of two models in a combined model. I want to make sure you fully understand the process as I described it for a fully reversed loading.nHere is the example of Solution Combination using Design Assessment. https://forum.ansys.com/discussion/comment/92431#Comment_92431nI will let the Fatigue experts answer the other questions.n

September 29, 2020 at 7:44 pmDaniel ShawAnsys EmployeeThe loading ratio is relationship between the minimum and maximum loading in a load cycle. It is often called the rratio (rratio = Smin/Smax). The FT does not use the actual loading. It just ratios the stresses. So, if the calculated stress is 500 MPa and the loading ratio 1 (fully reversing), the stress range is from 500 MPa to +500 MPa; and the alternating stress used by the FT is 1,000 MPa. If the loading ratio is 0.5, the stress range is from 250 MPa to +500 MPa; and the alternating stress is 125 MPa.nIf you have SN curves for different levels of mean stresses, you would typically specify the mean stress correction method to be ?Interpolate?. The FT would then use the appropriate SN curve for the calculated mean stress value (on a nodebynode basis). If the calculated mean stress fell between SN curves, the FT would interpolate to find the appropriate value.nYou typically specify a mathematical means stress correction (Goodman, Soderberg, etc.) when you only have a fully reversing SN curve (rratio= 1). The FT uses the alternating stress, the mean stress, and the appropriate formula from the specified method to calculate an equivalent alternating stress that can be used with the fully reversing SN curve. If you specify Soderberg with multiple SN curves, it will use the fully reversing SN curve to calculate the equivalent alternating stress.n

September 30, 2020 at 8:12 amRameez_ul_HaqSubscriberThank you for answering. ''If the loading ratio is 0.5, the stress range is from 250 MPa to +500 MPa; and the alternating stress is 125 MPa'', did you mean the alternating stress is 250 MPa, since 500  250 = 250. nWhere can I find the 'interpolate' choice, for mean stress correction? I mean I just see 'Mean stress curves' choice under OPTIONS tab, and then under 'Mean Stress Theory' option. Does selection of this choice means interpolation between the necessary mean stress curves?nIf the mean stress value doesn't fall between two of the known mean stress curves, then how does the ANSYS algorithm work?nWhen using a criterion like Soderberg or Goodman or Gerber, the value gotten from the equivalent Alternating stress, say 500 MPa, basically means the stress is alternating between 500 MPa and 500 MPa, since the loading is being transformed into fully reversed cycle, ALTHOUGH WE HAVE USED A LOADING RATIO instead of fully reversed cycle option? Am I correct, Sir?.n

September 30, 2020 at 3:56 pmDaniel ShawAnsys EmployeeSorry for the typo, but my mistake was with the fully reversing example. Stress range is the difference between the min and max stress experienced during the loading cycle. The alternating stress (aka stress amplitude) is the stress deviation around the mean stress experienced during the loading cycle. So, for the fully reversing case from 500 MPa to +500 MPa, the stress range is 1000 MPa. The mean stress is 0 MPa. The alternating stress is +/ 500 MPa around the 0 MPa mean. For the r=0.5 case, the stress range is 250 MPa (250 to 500). The mean stress is 375 MPa. The alternating stress is +/ 125 MPa around the 375 MPa mean.nTo invoke the mean stress interpolation option, specify the Mean Stress Theory to be ?Mean Stress Curves?. If the calculated mean stress lies directly on one of the mean stress SN curves (which is unlikely except for very simple models), the FT uses that curve. If the calculated mean stress lies between mean stress SN curves, the FT interpolates between the SN curves.nAny of the mathematical mean stress correction theories, use the mean and alternating stresses (along with either SY or UTS) to calculate an ?equivalent? alternating stress. The equivalent alternating stress is the alternating stress that will produce the same fatigue damage using the full reversing SN curve that the true alternating and mean stress would produce using a SN curve developed for that mean stress. They are mathematical methods to include the effect of mean stress while using the fully reversing SN curve. They were developed, and are used, because it is very common to only have the fully reversing SN curve for many materials.nAs an example, consider a loading cycle with a min stress of 0 MPa and a maximum stress is 500 MPa with the Goodman mean stress correction theory. The stress range is 500 MPa. The mean stress and the alternating stress are both 250 MPa. Per the Goodman theory (with an UTS of 1000 MPa), an equivalent alternating stress of 333 MPa along with the fully reversing SN curve would produce the same fatigue damage as the actual alternating stress of 250 MPa along with a 250 MPa mean SN curve.n

September 30, 2020 at 5:15 pmRameez_ul_HaqSubscriberThank you answering in detail, Sir.nFirst of all, you wrote, 'For the r=0.5 case, the stress range is 250 MPa (250 to 500). The mean stress is 375 MPa. The alternating stress is +/ 125 MPa around the 375 MPa mean'. But I would only be able to see the equivalent alternating stress right? Since this is the only option I see under the fatigue tool. Meaning, I donot know what the mean stress is, maximum value is, minimum value is, but I only know the amplitude stress for the loading ratio I provided i.e. 0.5. (But if we use the equational definition of the loading ratio and stress amplitude, then I think we will be able to find the above mentioned unknown parameters, by hand calculations).nCan you please tell the background of the mean stress curve? For a Ratio value of 0.5, we just have one curve where value of min stress / max stress is equal to 0.5. For the SN curve of R ratio of 0.5, we have stress amplitude against the number of cycles. Now, for the same stress amplitude, the mean stress can be different. It doesn't mean that a point on the SN curve of R ratio of 0.5 have a constant mean stress, although its alternating stress will be constant. For different points on the SN curve of R ratio of 0.5, the mean stress can be the same (or different), for each point having different alternating stress. Same is true for any value of R Ratio, it be 0.5, 1, 2, 2 etc. Now, consider the case of R ratio of 1 and 0.5, it can be possible for them to have the same mean stress but different amplitudes.nSo, I want to ask how does the ANSYS picks the closest R ratios' SN Curves? I mean depending on what? What if we have defined a loading ratio of 2 under the fatigue tool, but we donot have, for example for Aluminum Alloy, an R ratio value of 2 built into the engineering data sources.

October 2, 2020 at 7:26 pmDaniel ShawAnsys EmployeeThe FT outputs the equivalent alternating stress for the loading cycle, which is the stress used along with the SN curve to determine the allowable number of cycles. It does not output the intermediate stresses (maximum stress, minimum stress, and mean stress) used to calculate that equivalent alternating stress. You could use the equivalent alternating stress, loading ratio, and mean stress correction theory to back calculate the min, max, and mean stress for a loading cycle, but it is not clear why you want those values. nSN curves are developed by subjecting test specimens to a loading cycle. Most fatigue tests are conducted under fully reversing loading (r=.1). So, there is no mean stress for that loading cycle, and the SN curve developed from those tests represents the fatigue behavior with no mean stress. To include the effect of a mean stress in the CAE fatigue evaluation, you need to calculate an ?equivalent? alternating stress assuming one of the mean stress correction theories. However, SN tests can be conducted for loading cycles that produce a mean stress. Those SN curves represent the fatigue behavior for that actual level of mean stress. nThe FT calculates the min, max, mean, and alternating stress for the loading cycle, so it can calculate the rratio (Smin/Smax). If you instruct the FT to use ?Mean Stress Curves?, it calculates the rratio for the loading cycle (on a nodebynode basis) and uses the appropriate SN curve to calculate the allowable number of cycles. Consider a material with three SN curves: rratio = 1, ratio=.5, and ratio = 0. If the calculated rratio is exactly 0.0, 0.5, or 1.0, the FT just uses the appropriate SN to determine the allowable number of cycles. If the calculated rratio does not exactly match one of the SN curves, the FT interpolates between curves (loglog) to determine the allowable number of cycles. For example, if the rratio is r0.75, the FT would interpolate between the 0.5 and 1.0 curves to determine the allowable number of cycles. If the calculated rratio is larger or smaller than any of the defined curves, it uses the closest SN curve.nThe rratio is just the min stress divided by the max stress during a loading cycle, so clearly different stresses can produce the same rratio; 100 MPa to +100 MPa and 50 MPa to +50 MPa both have a ratio = 1.n

October 6, 2020 at 2:28 pmRameez_ul_HaqSubscriberHello there.nConsider this example. I conducted a 'Large Deflection On' analysis for a load of 100 N applied on a structure, now I have this analysis' solution and results, and I want the fatigue loadings to vary from 200 N to 50 N. I give it a scale factor of 2 (to represent 200 N of load), and a loading ratio of 0.25 (to represent 50 N) and conduct the fatigue tool evaluation. Does the FT do this in linear manner? I mean for the 2 scale factor, does the FT just doubles the equivalent stress at a specific location of the structure? Or there is a mathematical model behind it? Since doubling or decreasing to a factor of 0.5 of the current loading doesn't necessarily mean that the equivalent stress, at that specific location, will also do the same. How to incorporate a geometric nonlinearity to the FT?n

October 6, 2020 at 7:06 pmDaniel ShawAnsys EmployeeThe scale factor in the FT is a linear scale factor. Any nonlinear behavior must be captured directly in the FE results.nn

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