TS...

I THINK I know what You are talking about... 'NASA Astronautics Structures Manual'... but others may not.

PLEASE BE SPECIFIC. What NASA document number(S) are You inquiring about??? http://everyspec.com/NASA/NASA-General/NASA_TM-X-7... ????

BTW the ones I'm aware-of were published 'mid 1975'...

Regards, Wil Taylor
o Trust - But Verify!
o We believe to be true what we prefer to be true. [Unknown]
o For those who believe, no proof is required; for those who cannot believe, no proof is possible. [variation,Stuart Chase]
o Unfortunately, in science what You 'believe' is irrelevant. ["Orion", Homebuiltairplanes.com forum]

TM X-73305,6,7 ?

If you think you've an error post it here, at least for discussion.

another day in paradise, or is paradise one day closer ?

NASA TM X-73305, August 1975. Error is in Table B 5.1.4-1, statically indeterminate frame reaction forces, case 14 - fixed legs, linear increasing distributed load on a leg. Page is dated 12 Sept 1961. The X11 equation is

X11 = w/[60h²(d-b)] * [5hd⁴ - 3d⁵ - 20hdb³ - 12b⁴(d+h)]

Should be

X11 = w/[60h²(d-b)] * [5hd⁴ - 3d⁵ - 20hdb³ + 15b⁴(d+h) - 12b⁵]

All other equations for the case are correct.

A guy here found it when he went to put together a calculator for it, and checked it against a FEM to make sure he coded it right. Results didn't match, and we eventually came to the conclusion that the ASM was incorrect. We rederived it starting with the concentrated load in Case 4, made that a differential distributed load, and integrated. After severe algebraic flogging, managed to get everything into the form of the equations in the ASM, except for the changes to X11 noted above.

Again, not sure if anybody has previously identified this, or if any collection of errata exists for the ASM.

Also in that same case, the MF equation includes an X22 term that should be X12. But that's a little less tricky.

With luck this correction doesn't explain any failures.

T.S. ...

There is another [possible?] alternative contact for You to ask the same question... and where I downloaded my 3-volumes of the NASA ASM [and other interesting stuff]. Not promoting, only mentioning...

Abbott Aerospace
https://www.abbottaerospace.com/ [for contact link]
https://www.abbottaerospace.com/reference-data/nas...

Like I mentioned earlier... these NASA manuals were published in 1975. I suspect there has-to-be a sanity check on most of the documents contents... and a basic errata/change document... and/or unpublished updated ASMs for internal NASA use.

Regards, Wil Taylor
o Trust - But Verify!
o We believe to be true what we prefer to be true. [Unknown]
o For those who believe, no proof is required; for those who cannot believe, no proof is possible. [variation,Stuart Chase]
o Unfortunately, in science what You 'believe' is irrelevant. ["Orion", Homebuiltairplanes.com forum]

I sent an email to Abbott Aerospace, and haven't heard back.

While I'm at it, another error in the same table, case 1, pinned legs, concentrated load on the crossbar, shows H = 30ab/(2Lh(2K+3)), which is clearly an error, as it isn't dependent on the load, Q. And the fix is, simply, H = 3Qab/(2Lh(2K+3)).

FWIW, Abbot's spreadsheet makes that correction for case 1. https://www.abbottaerospace.com/downloads/aa-sm-02...

There is just about 0.0 chance of finding an errata doc for a NASA report. I have never seen one in decades in this business.

These documents were published by NASA, so maybe try contacting... NASA?

I got my copies from the NTRS. I know that for documents which are listed on the NTRS but not publicly visible, there is a form you can use to request access. I know from experience that they actually get back to you quickly. My first step would be to pose your question at the STI info desk here:
https://sti.nasa.gov/contact-us/#.YBsEg-hKiUk

However, these are already available and I see no errata associated.

So the alternative would be to contact someone at Marshall Spaceflight Center where they were written.

Looks like this guy might be who you want:
https://www.linkedin.com/in/larry-leopard-0b958b26...
https://www.nasa.gov/centers/marshall/news/news/re...

Or you could start with Marshall's social media contact and figure out who should field the question:
Marshall Space Flight Center
Jena Rowe
256-544-5022
jenalane.rowe@nasa.gov

General office of communications:
https://www.nasa.gov/centers/marshall/mediaresourc...

Finally, there is a note in the front matter of the manuals:
"This document has also been reviewed and approved for technical accuracy"
Although I can't imagine they were too thorough if they didn't find any errors since there are over 2000 pages of data. I think chances are low that there are any errata published.

On an interesting side note, the Marshall director at the time was Alex McCool. He was a pretty "cool" guy, you can see his signature of approval on the manual.
https://en.wikipedia.org/wiki/Alex_McCool

Sadly he passed away just recently, July 2020.

Keep em' Flying
//Fight Corrosion!

I know a couple of structures folks at NASA, I've asked a few of them. So far, no luck. Leopard might be worth a shot, maybe I'll reach out to him.

More errors in the table, still minor stuff. Case 7, pinned legs, linear increasing load on the crossbar, H has two separate equations,

H = 3/(2h) * (X3+X4)/(2K+3) = 3wc/(4Lh(2K+3)) * (dL - c^2/18 - d^2)

The first of these is missing a minus sign; should be

H = -3/(2h) * (X3+X4)/(2K+3) = 3wc/(4Lh(2K+3)) * (dL - c^2/18 - d^2)

Same case, the special case, the eqn for V is written as if the vertical forces are symmetric, which obviously isn't the case for asymmetric loading. The correct forces are

VA = wL/6, VF = wL/3

Hmm. All these errors are pretty sloppy work by NASA if they are true. Might require some more verification.

How specifically are you coming up with your revised formula for X11? You say you are taking a formula from Case 4, which is for a single point load, and treating this as a differential load of the distribution acting on a differential distance "a" from what I can tell. But which formula are you integrating? Case 4 doesn't have any formulas "X..", you just compute the reactions from load and geometric data. So are you integrating the equation for "H_e" then? Basically saying H_e is a function and it's variables are all differentials? If so, I'm confused how you are going about that to revise the formula for Case 14 X11 because Case 4 H_e would lead you to a function of two variables. The vertical position "a" is varying for a differential of a distributed load, and also, because it is a linear distribution, the value of Q is a variable. So you would need to be doing multivariable integration.

Also, note that Case 13 is identical, except the direction of the linear increase in load is the opposite. What happens if you change the loading in your FEM and compare to that? If you can't get your FEA to match a similar case, it might be a problem with your modelling, not the ASM.

I checked a couple other places... Roark does not have this specific case, although it does have the case where the load is distributed along the entire side of the frame. That is Table 8.2, page 207, Case 5g. The special case in the ASM Case 14 is exactly that.

However, AFFFL-TR-69-42 does in fact have this case, and reports all the same formulas as NASA. Maybe they got their data from the same source though.

I have a FEA script in Fortran that does beam, truss, and frame analysis. The FEA exactly matches Roark and the special case of Case 14 (by hand) where the distributed load extends the entire length of one side. So I know for sure that the special case is correct in the ASM. Although that doesn't help much because the terms you are suggesting need to be corrected drop out in the special case because b = c = 0.

Basically, I'm not convinced that it is actually an error without seeing your "severe algebraic flogging".


Keep em' Flying
//Fight Corrosion!

From the point load case, you take the load Q as dQ, at height x instead of a. Then, dQ varies between 0 at a and w at a+c, so you have

dQ = w * (x-a)/c * dx

Then, for example, the horizontal reaction at E force for the concentrated force,

HE = Qab/(2h^2) * [h/b - (h + b + K(b-a))/(h(K+2))]

For the differential form, change the a's to x's, and the b's to h-x, and you get

dHE = dQ * x(h-x)/(2h^2) * [h/(h-x) - (h + h-x + K(h-2x))/(h(K+2))]

Then plug in for dQ,

dHE = w/c * x(h-x)(x-a)/(2h^2) * [h/(h-x) - (h + h-x + K(h-2x))/(h(K+2))] dx

And integrate from x=a to a+c. Same for the vertical reaction force and moments. Integration is done symbolically in Mathcad, and unfortunately I don't have a copy of those I can post here. But they are transcribed in the attached PDF.

This gets you a set of ugly expressions, and from those, it's not obvious how to rearrange that into something like the ASM's form. But what I did was assume that the error was in X11 or X12, and solve the ASM equations for X11 and X12 in terms of the reaction forces. Then plug the ugly expressions from the integral into those, working by hand, and stuff starts to drop out, and you wind up with, X12 is correct, and the correction in my earlier post to X11. See attached.

I'm aware that this is prone to error, but I'm checking the raw result of the integral against a simple bar element FEM. The bar element FEM matches the integral solution to within a couple percent, where the as-written ASM solution is 25% off. And the simplified version matches the integral.

I've also been looking at the same FEM with concentrated loading and the special case loading, and comparing that with ASM and Roark. The FEM matches those just fine. Also took Roark's formulation for the distributed loading and was able to rearrange it to match the ASM's special case under the assumption that the legs are equal length and equal moment of inertia, and the whole frame is equal modulus.

So, I'm very sure that there is an error in the ASM table (which is maybe not surprising, given the sloppiness of the other errors), and that my result is correct. I invite you to compare a FEM of the general case to the ASM's result; it's a couple minute's work.

FWIW, I'm working on a similar error in case 8, which is a distributed loading along the crossbar, fixed legs. We've done the same integration technique, and confirmed it with a match to a FEM, but I haven't worked out the simplification yet.

• https://files.engineering.com/getfile.aspx?folder=1ab98fa4-5a69-4717-9898-2

OK, that is a bit more clear to me now. Although Q is also varying, it is a function of position, so you write it in terms of position and integrate in only one variable.

You say that you are calling the position of the differential load "x", and that dQ at "a" is zero, and dQ at "a + c" is w. This is still confusing to me. This would mean that the start point of the triangular distributed load is "a" and the end point is "a+c". However, in Case 14, "c" is defined as the start point.

I would've defined it:
X is the position of a differential load dQ, measured upward from point a to a final point c
dQ varies with X based on the slope of the distribution so, we have dQ = (w/(c-a))*x dx.
Then I follow everything until you substitute for dQ.

If you are defining the linear distributed load going from two points a to c, if you put in a value of (c-a) in for X in the dQ equation you defined, it should return the max value, W, but it doesn't.

Since you are integrating from x=a to x=a+c, I think you should be substituting ((w*x)/(c-a)) dx for dQ. Then integrate.

Maybe I'll make 2D FEM for the generic case this weekend, but TBH I think that might be overkill. There are a plethora of 2D frame analysis spreadhsheets and programs available on the web, many using the matrix force method described by Argyris.

I made a worksheet that performs the ASM calculation for Case 14, as given in the manual. I can confirm it is working correctly with generic inputs for the geometry because when I set b=c=0 it returns the same answers as the equations for the special case.

Then I made a duplicate where instead of the manual's equation for X11, I used your correction (that is the only change).

Then, I used this spreadhseet [https://www.cesdb.com/frame.html] with the exact same geometry and loading. Again, it matches the answer perfectly for the specialized case. But when I use the same load parameters I did with your corrected version of the ASM calculations, the answers don't come out the same.

These are my parameters for everything:
frame height = 10 units
frame width = 5 units
load goes from 2.5 units to 7.5 units from 0 to 50.
assumed cross section of the frame is 1x1 and the second moment of areas based on bh^3/12.

With the ASM formulas (including your correction of X11) I get horizontal reactions of 36.91 and 88.086 for equilibrium with the 125 distributed.

With the spreadsheet, the horizontal reactions come out to 92.53 and 32.47. Try using a 2D frame too from the web and see if it matches your calcs.

Instead of using Case 4 and adapting it, I'd think it would be better to just get the reactions from scratch using the matrix method
https://reports.aerade.cranfield.ac.uk/handle/1826...

Keep em' Flying
//Fight Corrosion!

Yeah, I apologize, I was going off of memory and getting the geometry confused with another case.

I think you may have entered the expressions into your spreadsheet incorrectly. When I set up your case in my spreadsheet, I match the result from cesdb.

I'm not going to claim that the integral method is the cleanest way to do it. One of my colleagues did it and had Mathcad spit out the result, so that was my starting point.

Ok, finished digging into these. The complete list:

Case 1: H = 30ab/(2Lh(2K+3)) should be 3Qab/(2Lh(2K+3))
Case 7: X3 = -wc/(2L) * [d³/L + c²/9 + 51c³/(810L) + c²b/(6L) - d²] should be X3 = -wc/(2L) * [d³/L + c²/9 + 51c³/(810L) - c²b/(6L) - d²] (Plus sign on the c²b/(6L) term should be minus. This affects Case 8, as well.)
Case 14: X11 = w/[60h²(d-b)] * [5hd⁴ - 3d⁵ - 20hdb³ - 12b⁴(d+h)] should be X11 = w/[60h²(d-b)] * [5hd⁴ - 3d⁵ - 20hdb³ - 15b⁴(d+h) - 12b⁵]
Also Case 14: MF = (3K+1)/[2(6K+1)] * [w(2a²-ac-c²)/6 - X11] - X22/2 * [1/(K+2) - 3K/(6K+1)] should be MF = (3K+1)/[2(6K+1)] * [w(2a²-ac-c²)/6 - X11] - X12/2 * [1/(K+2) - 3K/(6K+1)]

Case 1 is pretty straightforward; obviously it should be dependent on the applied load Q, and it's a simple typo of 0 for Q.

Case 14, I'm not sure exactly how that first error occurs, but see above for derivation; the second is an obvious typo.

Case 7/8, the easy way to see that is, Case 7 gives you two expressions for H, one with X3 and X4, and one without:

H = 3/(2h) * (X3 + X4)/(2K + 3) = 3wc/[4Lh(2K+3)] * (dL - c^2/18 - d^2)
X3 = -wc/(2L) * [d³/L + c²/9 + 51c³/(810L) + c²b/(6L) - d²]
X4 = wc/(2L) * [d³/L + c²/18 + 51c³/(810L) - c²b/(6L) - 2d² + dL]

And if you take the first expression for H, and add X3 and X4 together, the c²b/6L terms don't cancel like they have in the second expression. From a FEM, the second expression appears correct, so I suspected the error to be with the sign on one of the c²b/(6L) terms. From the Case 8 reaction forces, that appeared to be on X3. Working through a similar process of integration of the concentrated loads in Cases 1 and 2 to get analytical solutions, changing the sign on the c²b/(6L) term in X3 does, in fact, correct the error. It is, once again, a bunch of algebra.

Results are verified by comparison to a bar model FEM. I also compared to Roark's frame solutions (Table 4 in the 5th ed) (Case 1, you've got the general solution; Case 7 and 8 you've got the special case of the load going the full width, which is a limited but still useful comparison, because you can't get to the special cases in the ASM from the X3, X4 formulation; Case 14 is only the special case of the load going the full height, which isn't an interesting comparison because the error cancels out for the special case).

Unfortunately for the record here, I documented that approach electronically and not on paper, so I'm not going to post anything more than the result here.

FWIW, all I've been able to find as far as errata in the ASM goes is a memo from MSFC from I believe 1988, ED24-88-48, identifying an error in Section B 4.1.4, page 27, eq 1; the kd/2 term should be 2kd/3; that carries into eq 2, but the memo leaves it as an exercise for the reader.