MLife/Interpretation of DEL and LUlt

Dear Jason,
I have few questions regarding the interpretation of fatigue load outputs from mlife. The approach ive used is as follows (the cross section of interest is root),
1). Since the LUlt of the cross section of interest isn’t available, I’ve ran the analysis for different LUlt . Therefore from the analysis we can get the ~LUlt required for the 20 year life time. (please see the image

2). The damage fraction is also checked for the same LUlt, to ensure that the damage is < 0.75-1 (please see the image

3). Then the approximate DEL for the LUlt is obtained from the DEL for various LUlt graph.
(please see the image

If my above interpretation is correct, now since we know the ~LUlt, Ixx or Iyy for the cross section of interest and the y then from euler bernoulli (LUlt*Ixx)/y = M. can I compare this Moment and the Largest moment obtained from FAST (EWM50)?. And what does it mean if it’s less than or greater than the EWM50 moment. or can I use this DEL to check the stress to ensure that this cross section can with stand this DEL.

Thank you for your time,
Subramani Anbazhagan

Dear Subramani,

Your plots make sense to me, except for the damage-equivalent load (DEL) for RootFy, which I would expect to asymptotically approach a fixed value for large LUlt, but the plot seems to drop a bit at LUlt = 1000 kN.

I’m not sure I understand your equation LUltIxx/y = M. I think you are trying to use the flexure formula sigma = My/Ixx, but you’ve exchanged sigma for LUlt, where sigma is the stress. LUlt is a load, not a stress. Using an LUlt lower than the load derived from an extreme load case would imply failure.

Ideally, LUlt would be based on a strength (e.g., FEA) analysis of the cross section of the component in question, as discussed in our FAQ: nwtc.nrel.gov/FAQ#TypeLMF. For a simple beam and pure bending, for example, LUlt could be derived from the flexure formula above: LUlt = sigma_yield*Ixx/y_max, where sigma_yield is the yield strength of the material (perhaps reduced by a safety factor) and y_max locates the point in the cross section of maximum stress.

I hope that helps.

Best regards,

Dear Jason,
Thank you for the response. it really helped a lot. I’ll look into it for the 1000 kN Lult case. I have few more question’s,

1). Partial Safety Factors input which i provide for the input files (mlife Input files). There are four PSF’s (PSF1…PSF4). From the IEC document, for the fatigue damage estimation we need to use three PSF’s ( for loads, materials and consequences of failure). I was wondering what is the fourth PSF input? and also i would like to know whether its in same order i mentioned above?.

2). And I’m still confused with the LUlt, so LUlt’ll always be in terms of moments right or it depends on the type of channel which we are looking at?. I’m aware that it’s the ultimate load that the cross section can with stand, but if i’m looking at force channel, would the LUlt be in terms of force?

Thank you for your time.
Subramani Anbazhagan

Dear Subramani,

There are four partial safety factors in the MLife input file (PSF1, PSF2, …, PSF4), but they don’t correspond to specific loads, materials, or consequences of failure. Instead, they are just placeholders for user-specified values (e.g. 1.35 for normal extreme, 1.1 for abnormal extreme, 1.0 for fatigue. etc.). The actual PSF used will depend on the setting of PSF_Type in the Input-Data Layout section of the MLife input file e.g. when PSF_Type = 2, the value specified for PSF2 will be used.

LUlt corresponds with the channel e.g. for a moment channel, LUlt is a moment; for a force channel, LUlt is a force.

The flexure formula (sigma = M*y/Ixx) only applies to a simple beam undergoing pure bending. For a simple rod undergoing axial loading, the formula would be sigma = F/A. A more general cross-sectional analysis is required for a more complicated multi-component stress state (axial, shear, bending, etc.), especially for a nonaxisymmetric cross sections composed of an anisotropic material.

Best regards,

Dear Jason,
Once again thank you for the response. I’m afraid I have to add few more questions,
1). The DEL is 47 kN.m (for the rootMyb1), TypeLMF is 2 kN.m (unknown, set to Aggregate Mean), then the DEL oscillates between 25.5 kN.m and -21.5 kN.m. is my understanding of DEL is right?. But the largest moment in this channel is 7 kN.m (from EWM50 wind), but i didn’t quite understand why the DEL is far bigger than the ultimate moment of the channel?.

2). And can I use this DEL above the TypeLMF (i.e., 25.5 kN.m) to convert it back to stresses using flexure equation and compare it with the material strength at the 1E7 cycles for calculating the fatigue safety factor for this channel?. If not please provide me some idea about it.

Thank you very much for your time.
Regards,
Subramani Anbazhagan

Dear Subramani,

Yes, a DEL of 47 kNm with a mean of 2 kNm results in a sinusoidal moment oscillation between -21.5 kNm and 25.5 kNm. I’m not sure where you might have made a mistake, but 7 kNm does not sound like the correct moment maximum for this range. Instead, I would expect that the maximum of the moment time history would be beyond this range.

Yes, you should be able to convert the fatigue load to stress, just as you would an ultimate load.

Best regards,

Dear Jason,
Thank you very much for your responses. I apologize that i have to add few more questions.
1). The amplitude of the DEL from zero-mean would be higher than the amplitude of DEL from constant fixed mean, since the zero mean has higher cycles to failure than the fixed mean, so in order to attain the same damage as the stochastic wind, the amplitude of zero mean needs to be higher. Is my understanding of DEL is right?

2). Based on the previous post, I’ve compared the extreme bending moments with hand-calc by assuming the blade to be flat on facing the extreme wind. Now I suppose it’s in the acceptable range, the DEL (Fixed mean/upper asymptote) of the flap-wise root bending moment is 4.21kN.m and the Extreme root bending moment is 7.5 kN.m. But this time DEL for the other span stations (please see the image) after 3.5m crosses the extreme bending moments. I’m wondering whether it is because of a wrong setting that i use in MLife or something relating to the design aspect of the structure. please provide me suggestions about it.


once again thank you very much for your time. Have a nice day ahead.
Regards,
Subramani Anbazhagan.

Dear Subramani,

Yes, the amplitude of the DEL from zero-mean would be higher than the amplitude of the DEL from constant fixed mean. This follows from the Goodman correction (Eq. 3 in the MLife Theory Manual) with L_RF = L_R0 and L_MF = 0, whereby the term in parentheses is greater than unity.

I’m not sure I understand your plot because it shows the DEL from zero-mean to be lower than the DEL from fixed mean, in contradiction to the above point. Another oddity is that you are comparing the DEL directly with the extreme load. As we discussed previously, a given DEL and L_MF will result in a sinusoidal load oscillating between L_MF-DEL/2 and L_MF+DEL/2. So, it would probably make more sense to compare the extreme load to L_MF+DEL/2. Furthermore, your case involving a flat blade facing the extreme wind will result in a large load, but you haven’t stated how close this load is to the ultimate load of the cross section L_Ult, or to the highest load found from all the load-case simulations that were used to calculate the DEL.

Best regards,

Dear Jason,
My apologies for getting back late. Thank you very much for your response.
Yes i agree with your point, it’ll make more sense, if i compare the peak of the DEL of both fixed mean and zero mean. Indeed what I’ve plotted is the mean + del/2 for fixed mean and its wrong in the zero mean. (i.e., For the 0.8m radial location the DEL_fixed mean and fixed mean is 2.48 kNm and 2.54kN.m , therefore the peak is 3.78 kN.m and DEL_zero mean is 2.98kN.m, the peak is 1.46kN.m). spn_DEL.png

In case of the root, DEL_fixed mean peak is 4.685 kN.m and the L_Ult of the section is 29.5kN.m, bending moment for EWM50 is 9.5kN.m, is my approach is right?. if in case its right, what could be the reason for the DEL to go above the extreme bending moment.

Thank you very much for your time.
Regards,
Subramani Anbazhagan

Dear Subramani,

Yes, your approach sounds correct now.

Regarding why LMF+DEL/2 is more than the load from EWM50 (at radial locations larger than 3.5 m) makes me question whether the EWM50 case actually produces the highest load. What is the highest load (as a function of radial location) from all of the load-case simulations that were used to calculate the DEL?

Best regards,

Dear Jason,
Thank you for the response. I didn’t ran the analysis for DLC cases other than EWM50 and EWM01, If it could make much differences to the flap-wise moments, I might need to focus my efforts on it.

Thank you very much for your time.
Regards,
Subramani Anbazhagan

Dear Jason,

I have query regarding the calculation of LUlt in MLife.

I have calculated the LUlt by calculating the maximum loads or moments(Interface and mudline) based on 10 load cases in MExtrermes and multiplying this value by factor of 5 to get the LUlt. Is this the correct way of finding out the LUlt, if not, could you please advice on this?

Thanks,
Satish J

Dear Satish,

LUlt has been discussed many times on our forum, e.g. see:

http://forums.nrel.gov/t/mlife-fatigue-del-calculation/661/29
http://forums.nrel.gov/t/mlife-user-defined-distribution/1970/15

Best regards,