Fatigue-Life Prediction of Mechanical Element by Using the Weibull Distribution
Round 1
Reviewer 1 Report
This work summarized adopted the known methods (Weibull method) and applied them to determine the expected failure times. It was well written and all the methods are clearly described. Two major comments were listed for the authors to improve the manuscript for final publication.
- The background search and introduction need to be consummated. There are tons of work that tried to estimate the Ni through different statistical methods. Please go through more literature and review them in your manuscript
- In figure 4 and 5, this works shows the stress concentration on the end point of the beam. It is known that the FEA model end point stress concentration max value is highly related to mesh size and strategy. In this work, the case was very simply but still has the potential that different mesh sizes will change the results. Please use a finer mesh size and run the model again to confirm if you can have the same result or not. If the stress value was different, then the FEA data cannot support your method.
Author Response
Response to Reviewer (1) Comments
Point 1: The background search and introduction need to be consummated. There are tons of work that tried to estimate the Ni through different statistical methods. Please go through more literature and review them in your manuscript.
Response 1: You are right, there are a lot of methods to statistically estimate Ni. Mainly, as mentioned in the three added references they are based on the stress, strain, and crack size. However, the advantage of our proposed method is that it completely reproduces the principal stresses and since the random behavior of the derived Ni values is around them our proposed method is efficient. References were added in the introduction section.
Point 2: In figure 4 and 5, this works shows the stress concentration on the end point of the beam. It is known that the FEA model end point stress concentration max value is highly related to mesh size and strategy. In this work, the case was very simply but still has the potential that different mesh sizes will change the results. Please use a finer mesh size and run the model again to confirm if you can have the same result or not. If the stress value was different, then the FEA data cannot support your method.
Response 2: The method’s stress inputs are the principal stresses. Therefore, if they are affected (e.g. mesh size, etc.) the newly affected principal stresses values will be the method’s input. It is to say the efficiency of the stress method remains. Additionally, observe the random behavior of Ni is performed around the N value given by the Basquin’s equation. Thus, the efficiency of Ni depends on the efficiency on which the Se endurance limit was determined and on the used failure theory. Here Se was determined in the static analysis by considering the modifying factors, and the equivalent stress was calculated by using the Elliptical failure theory. It is to say if a reader determines Se and equivalent stress by using another failure theory or modified factor combination, only the corresponding Ni value will change, but the method’s efficiency remains.
However, a finer mesh size was used, the stress values obtained were different by decimal.
- Current stress values are: = 75mpa and = 184.8mpa for a mesh of 393 nodes and 40 elements.
- New stress values are: = 77mpa and = 184.4mpa for a mesh of 1343 nodes and 160 elements.
Thus, since by using a finer mesh size the change in the principal stresses values were not significant, we conclude the FEA results support our method
Reviewer 2 Report
Dear authors thank You for this interesting paper. The authors should work on some details which would improve the quality of the paper in order to be accepted:
- The authors state that there are no methods which would predict the actual fatigue life in number of cycles for mechanical elements for the case of random loading. This is important especially for the equivalent stress method also taking into account the mean stress effect, but we can see a lot of fatigue life estimation methods for random loading like e.g:
DOI: 10.1520/MPC20150049 - The Weibull probability density is one of the extreme cases while predicting fatigue life for random loading due to the fact that such a distribution is very rarely observed for real life machine components, did the authors thought about confronting their methodology with other probability densities models such as the Rayleigh, Zhao-Baker or Dirlik model, which as many scientists demonstrated are usually more practical in terms of calculations
- The paper should be double checked in terms of the font size and the spacing between the lines as some paragraphs seem to be different.
- Page 11 line 322 :"Using the Weibull stress family results, the stress random behavior can be determined. The procedure is as follows:" The authors should think about working on the structure as after the colon we get a new paragraph and not the pointed out procedure.
- The authors should show a plot of the obtained probability densities in regards of amplitude distribution in order to compare the proposed theoretical model to the calculation results.
- The conclusions should be more concise.
Author Response
Response to Reviewer (2) Comments
Point 1: The authors state that there are no methods which would predict the actual fatigue life in number of cycles for mechanical elements for the case of random loading. This is important especially for the equivalent stress method also taking into account the mean stress effect, but we can see a lot of fatigue life estimation methods for random loading like e.g: DOI: 10.1520/MPC20150049
Response 1: Because the validation of an empirical model occurs when it reproduces its inputs, then since these inputs are the σ1=491.75 and σ2=184.8 values, and due to by applying the proposed method we always can reproduce and then no comparison with others distributions is needed. This fact that and are reproduced can be seeing in Table 4 between rows 5 and 6, where the σ1=491.75 and σ2=184.8 values were reproduced. Moreover, observe because they were reproduced then the corresponding mean stress σm= (σ1 + σ2)/2 = 338.275 and the amplitude stress σa= (σ1 - σ2)/2 = 153.475 values are also reproduced. (Please see also comment 2).
Point 2: The Weibull probability density is one of the extreme cases while predicting fatigue life for random loading due to the fact that such a distribution is very rarely observed for real life machine components, did the authors thought about confronting their methodology with other probability densities models such as the Rayleigh, Zhao-Baker or Dirlik model, which as many scientists demonstrated are usually more practical in terms of calculations.
Response 2: Because the Weibull distribution is completely defined to any σi or Ni values and it only depends on the β and η values, then because as demonstrated in Table 4, the estimated Weibull β and η parameters always reproduce σ1 and σ2, the Weibull distribution is the best distribution we can use to analyze fatigue data. Also notice that due to failure occurs at the fracture limit in the stress-strain curve then the fatigue fails must be modeled by an extreme distribution. Therefore, for deeper details of the Weibull properties and its adequacy to model fatigue data, on page 3, above sec. 2.1, readers are invited to read chapter 2 of reference [11].
Point 3: The paper should be double checked in terms of the font size and the spacing between the lines as some paragraphs seem to be different.
Response 3: It was corrected.
Point 4: Page 11 line 322 :"Using the Weibull stress family results, the stress random behavior can be determined. The procedure is as follows:" The authors should think about working on the structure as after the colon we get a new paragraph and not the pointed out procedure.
Response 4: This was corrected, now it is "And because these parameters only depends on the applied σ1 and σ2 values, then by performing the Weibull analysis their random behavior can be determined as follows".
Point 5: The authors should show a plot of the obtained probability densities in regards of amplitude distribution in order to compare the proposed theoretical model to the calculation results.
Response 5: In Weibull analysis, the amplitude is given by the η=301.4555 and µ=338.275 parameters as Sqr(338.275^2-301.455469^2)^0.5 = (491.75-184.8)/2 = 153.75. Therefore, because µ is unique (it comes from the diagonal element of the stress matrix which is invariant), and from Eq.(9b), η only depends on σ1 and σ2, then once σ1 and σ2 are known both µ and η are unique, implying the theorical model is correct. And consequently, no graph is necessary. However, the related pdf is included in the file attached.
Point 6: The conclusions should be more concise.
Response 6: The conclusions have been updated, now they are specific.
Author Response File: 
Author Response.docx
Reviewer 3 Report
Presented work is valuable and can be interested in others but have major shortcomings. The authors state that: it is not possible to determine the expected failure times (cycles to failure) (Ni) values of a mechanical element by applying the actual failure theories (Goodman, Gerber, etc.). It must be explained that the fatigue limit is associated with the phenomenon that crack nucleation is arrested by the first grain boundary or a dominant microstructural barrier [17]. By this definition, the crack can’t propagate below the fatigue limit. For very high cycle fatigue (e.g. cycles over 107) it can be found other S-N curve definitions see:
Marines, I.; Bin, X.; Bathias, C. An understanding of very high cycle fatigue of metals. Int. J. Fatigue 2003, 25, 1101–1107.
Sakai, T.; Lian, B.; Takeda, M.; Shiozawa, K.; Oguma, N.; Ochi, Y.; Nakajima, M.; Nakamura, T. Statistical duplex S-N characteristics of high carbon chromium bearing steel in rotating bending in very high cycle regime. Int. J. Fatigue 2010, 32, 497–504.
Pyttel, B.; Schwerdt, D.; Berger, C. Very high cycle fatigue - Is there a fatigue limit? Int. J. Fatigue 2011, 33, 49–58.
It should be explained why don’t use methods from other researches.
Particular comments are the following:
Could the authors explain the difference between stress-life analysis and fatigue life analysis? It is mentioned in lines 79 and 80.
Line 33 Please define σmax and σmin.
Line 35 Please give for reference [5,6] the number of pages you have been used.
Reference [8] Wrong title of the paper.
Line 136-138 There are symbols μy and μx. Is it a mistake?
Eq. (8) There is a constant 0.99176. But in eq. (46) I haven’t found that constant. Could the Authors explain it?
Line 137 Authors selected 21 samples for testing. It should be supported by standard requirements like ASTM E-739-91 or ISO-12107. Or by other researches. You can find it in reference [17] page 109.
Eq. (22) What means Ka, Kb, Kc, Kd, Ke, and Kf?
Line 266 It must be given the type of element that was used.
Table 1 The minimum and maximum deformation of the element is 3.06 mm and 8.15mm, respectively. But inline 255 are given 3 mm and 8 mm for minimum and maximum deflection, respectively. The authors should explain the difference. You can use more elements for FEA, then you should be error smaller.
Line 299 It is defined material AISI 4340 OQT 1300, but inline 234 is AISI 4340. Is it the same material?
Table 4 σmax and σmin are estimated by eq. (17). It is not understood. The stress of element is calculated from applied load F (Eq. (1)). Can you explain it?
Comments for author File: 
Comments.pdf
Author Response
Response to Reviewer (3) Comments
Presented work is valuable and can be interested in others but have major shortcomings. The authors state that: it is not possible to determine the expected failure times (cycles to failure) (Ni) values of a mechanical element by applying the actual failure theories (Goodman, Gerber, etc.). It must be explained that the fatigue limit is associated with the phenomenon that crack nucleation is arrested by the first grain boundary or a dominant microstructural barrier [17]. By this definition, the crack can’t propagate below the fatigue limit. For very high cycle fatigue (e.g. cycles over 107) it can be found other S-N curve definitions see:
Marines, I.; Bin, X.; Bathias, C. An understanding of very high cycle fatigue of metals. Int. J. Fatigue 2003, 25, 1101–1107.
Sakai, T.; Lian, B.; Takeda, M.; Shiozawa, K.; Oguma, N.; Ochi, Y.; Nakajima, M.; Nakamura, T. Statistical duplex S-N characteristics of high carbon chromium bearing steel in rotating bending in very high cycle regime. Int. J. Fatigue 2010, 32, 497–504.
Pyttel, B.; Schwerdt, D.; Berger, C. Very high cycle fatigue - Is there a fatigue limit? Int. J. Fatigue 2011, 33, 49–58.
It should be explained why don’t use methods from other researches.
Response: First, analyzing the fatigue phenomenon as the repeated application of constant stress, which due to the repetition of the stress generates deformation (∈=∈e+∈p) due to the loss of resistance of the material (decrease in the area), then as the stress applied is constant and is given by the values σ1 and σ2, then given that in our method the Weibull distribution completely reproduces them (see Table 4 between row 5 and 6, where σ1= 491.75 and σ2= 184.8 are reproduced), the method is efficient in reproducing this constant stress. With reference to the deformation generated until fracture, the method considers it in the Ni value of the S-N material’s curve. Although the deformation is not analyzed by itself because the failure time represented by Ni already contains its effect and since in our method the input related to the deformation Δ∈=(2σ'f/E)*(2N)^b is Ni, thus our method is only as efficient as the value of Ni has been determined. That is, the method is efficient to determine the random behavior of Ni, the critical thing is to determine the correct value of Ni. Due to it, other methods were not used.
Point 1: Could the authors explain the difference between stress-life analysis and fatigue life analysis? It is mentioned in lines 79 and 80.
Response 1: The stress life method determines the element’s life based on the applied stresses in our case they are σ1=491.75mpa and σ2=184.8mpa. for instance, from Table 5 observe that any element in column σ2i, say 66.35155 by using it in the reliability Eq.(7) with η=301.455469 the reliability given by the stress life method is 0.96729. Similarly, in the fatigue life analysis represented in Table 5 by Ni, by using any element, say 273683422.9 in Eq. (7) with η =1243427849 the reliability given by the fatigue life analysis is 0.96729. Observe that in our method both analyses are given, and they converge to the same reliability index. Finally, the above analysis can be observed from the fatigue accumulated damage formulation (ni/Nif), where ni represents the number of applied constant stress and Nif represent the fatigue life.
Point 2: Line 33 Please define σmax and σmin.
Response 2: They represent the Maximal principal stress σmax and Minimal principal stress σmin. To avoid confusion now in the article they are σ1 and σ2.
Point 3: Line 35 Please give for reference [5,6] the number of pages you have been used.
Response 3: For [5] pages 185-186, 188, and for [6] pages 313-316. The article will be updated accordingly.
Point 4: Reference [8] Wrong title of the paper.
Response 4: It was corrected.
Point 5: Line 136-138 There are symbols μy and μx. Is it a mistake?
Response 5: No, they are correct. The mean μy is stated per Eq.(15) as the mean of the Vector Y. And by Eq.(9b) μx represents the natural logarithm of the square root of the principal stresses σ1 and σ2
Point 6: Eq. (8) There is a constant 0.99176. But in eq. (46) I haven’t found that constant. Could the Authors explain it?
Response 6: This is because in Eq. 46 in [21] the value of β shown, only represents the initial value which should be iterated following the method of section 4.1 to finally determine the value of β that completely reproduces the values σ1 and σ2. Thus, the given constant 0.9921 in this article determines the value of β that reproduces the σ1 and σ2 values directly.
Point 7: Line 137 Authors selected 21 samples for testing. It should be supported by standard requirements like ASTM E-739-91 or ISO-12107. Or by other researches. You can find it in reference [17] page 109.
Response 7: This (n = 21) is not supported by following a standard, it is simply to estimate the parameters according to a desired reliability confidence level, in this case it was 95.35%.
Point 8: Eq. (22) What means Ka, Kb, Kc, Kd, Ke, and Kf?
Response 8: They are endurance limit Se modifying factors included in the Marin equation Eq.(22).
where Ka = surface condition modification factor
Kb = size modification factor
Kc = load modification factor
Kd = temperature modification factor
Ke = reliability factor
Kf = miscellaneous-effects modification factor
Point 9: Line 266 It must be given the type of element that was used.
Response 9: This was included: Finite element model uses the first-order element.
Point 10: Table 1 The minimum and maximum deformation of the element is 3.06 mm and 8.15mm, respectively. But inline 255 are given 3 mm and 8 mm for minimum and maximum deflection, respectively. The authors should explain the difference. You can use more elements for FEA, then you should be error smaller.
Response 10: The difference is due the mesh, after a finer mesh was used the deformation values were 2.99 mm and 8.02 mm respectively.
Point 11: Line 299 It is defined material AISI 4340 OQT 1300, but inline 234 is AISI 4340. Is it the same material?
Response 11: Yes, it is the same. The article was updated to include “OQT 1300” to line 234.
Point 12: Table 4 σmax and σmin are estimated by eq. (17). It is not understood. The stress of element is calculated from applied load F (Eq. (1)). Can you explain it?
Response 12: In Table 5 the values given in the column is referred to the minimum material strength that is required by the applied stress. For instance, in the first row of Table 5 by applying the stress, the required material strength is 66.35155, thus from Eq.(7) the desire reliability will be 0.96729. similarly, to the data of the column represent the Weibull scale parameter that the element must present to withstand a constant stress value of 301.4555. For instance, in Eq.(7) if we use t=301.4555 with η=1308.124 the reliability is 0.96729. Generalizing as you mentioned, the columns do not represent stress, they represent the expected value of minimum strength that the material should have in order to present the desired reliability.
Round 2
Reviewer 1 Report
Both shortcomings about literature review and simulation data validation were not fundamentally improved.
Author Response
Point 1: The background search and introduction need to be consummated. There are tons of work that tried to estimate the Ni through different statistical methods. Please go through more literature and review them in your manuscript.
Response 1 (R2): The inputs for the proposed method are σ1 and σ2, σeq, and Neq values and because the proposed method is original and not a modification of an existing one since the method completely reproduces the input values (see Table 4 and 5), in addition, determines the randomness of the cycles around the value N_eq (see table 5). So no further literature review was considered necessary. However, 3 recent references were added where the proposed statistical methods do not reproduce the input values or ensure the randomization of Ni.
Point 2: In figure 4 and 5, this works shows the stress concentration on the end point of the beam. It is known that the FEA model end point stress concentration max value is highly related to mesh size and strategy. In this work, the case was very simply but still has the potential that different mesh sizes will change the results. Please use a finer mesh size and run the model again to confirm if you can have the same result or not. If the stress value was different, then the FEA data cannot support your method.
Response 2 (R2): The simulation with a finer mesh was executed, the variation in the stresses values obtained were not significant. It's important to mention the following:
1- The FEA simulation was not carried out with the objective of validating the proposed method, but was carried out simply to cover the 2 common ways to obtain principal stresses. Once the FEA analysis was performed with a finer mesh, the results of the principal stresses did not show significant changes.
2- That the efficiency of the proposed method does not depend on the efficiency of the simulation since it only takes as an input value the principal stresses that represent the application, regardless of how they were calculated (in this case the principal stresses were obtained through the static analysis and finite element analysis), finding no significant difference between the values obtained.
Reviewer 2 Report
The authors have corrected the paper according to the reviewers amendmends, thats why i recommend to accept the paper in the present form.
Author Response
No suggestions were made nor corrections were required.
Reviewer 3 Report
All the comments have been addressed in this revision, with no further comments.
Author Response
No suggestions were made nor corrections were required.
