Analysis of Statically Indeterminate, yet very simple, Bracket 10

[nargonaut7]

Greetings everybody,

I am sorry that my question has already been presented in this forum, but the previous answers, at least those two that I have been able to find here, are not adequate, as I explain later.

My situation, as shown below, is that of the second-simplest possible structure: a two member-frame attached to a wall (this is the second-simplest structure because the two absolute simplest structures have only one member, which is either a cantilever beam or a single-standing column, with their supporting wall or floor, respectively; by the way, I am writing all this not with the purpose of giving a lecture but only to make my point as clearly as I can). So, it really bothers me that excellent books such as Roark’s “Formulas for Stress and Strain”, “Structural Engineering Handbook” by Mahamid and others, and “Structural Engineering Formulas” by Mikhelson and Hicks, these three books include formulas for much more complicated frames, when this simple bracket support I am showing here, which is used in industrial plants all over the world, is completely ignored. Neither this two-member frame is shown as an example in multiple books where I reviewed the Three-Moment Equation, the Moment Distribution Method, Castigliano Theorem, and the Slope-Deflection Method, to verify if I could apply any of them to my situation. A severe disappointment I got was with the Slope-Deflection Method, when the determinant of the 4 x 4 matrix of the coefficients of the variables, calculated with the particular dimensions and the weight “P” of my application (none of these parameters are needed to be indicated here) turned out to be zero. Another disappointment was when I used Roark’s Tables 8.8 and 8.9, to combine axial and transversal deformations using the fact that the deformations of both members at the joint “C” are the same. By trial and error, I found multiple combinations of the three internal reactions (two forces and one moment) at that joint that make those two vertical deformations equal, so that didn’t work either.

I thought I was close to calculating the reactions in this bracket, when I found the “Air Force Stress Analysis Manual”. Its 579-page PDF file can be easily downloaded for free by searching in internet with those words. The reactions and moments I calculated for a triangular frame (Table 5-5) with all three joints fixed and with an external moment applied (Case 8) satisfied all the Statics summation of forces, but this didn’t happen with an external load (Case 9), so I couldn’t combine those two cases to simulate my bracket, as was my plan. That Case 9 didn’t work even when I tried to solve the exact situation they show, which is that both members are inclined and that the load is applied between supports, not in the overhang as in my situation. So, when I couldn’t even solve the standard problem shown in the diagram of Case 9, plus my failures calculating with Roark and Slope-Deflection, as described earlier, that was when I realized that I need to ask for help. But, of course, these failures could be also because I made a numerical error.

If somebody wants to verify these formulas, or any other method, you can assume your own load, dimensions, and inertias for those two members, and check if the resulting reactions you get make all the Statics summations equal to zero.

Back to the previous answers, I have seen two of them, I want to mention this to save somebody’s time in replying. The first one is the advice to use an FEA software, which of course I will use, but I always want to see a manual calculation of forces and stresses, at least for simple cases like this bracket support, to be included in a design report together with software results. The second answer is the advice to assume all three joints as pinned. This simplification reduces the problem from statically indeterminate to determinate, and I already did that, then it becomes really simple, the maximum moment in AD is P times the distance CD, the same as if the joint “C” were a fixed support and then we only consider the overhang, I found that result surprising. This quick and simple approximation can be useful in many situations, we can always use a higher safety factor to cover us if the actual moments and stresses are higher than the calculation, but I am beyond this point, it has become a matter of honor for me to find the method or formulas to calculate all nine reactions with all three joints welded as accurately as practical. The caveat of being practical is important so, to clarify my position, the extremely long formulas of the Air Force manual and of Roark are fine with me.

Well, I will appreciate very much any help, and I hope somebody will find useful the four references I have included in this post. And sorry for writing so much.

 

[rb1957]

in 8, wouldn't Ha = -Hb ?

9 is showing that the triangular truss reacts the applied load as an ideal truss ... the inclined member will react the applied load, and the horizontal member carries the horizontal component on the inclined member load.

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

 

[Settingsun]

Celt83,
Yeah, you wouldn't be able to use my method when the beams aren't slender (like what you analysed in Mastan). Using your section properties and considering the section as just a pair of flanges (very efficient section), the depth is already deeper than your 6" overhang. Any real section would be deeper again to achieve that I value with that area.

 

[BAretired]

in 8, wouldn't Ha = -Hb ?

That seems right to me. Otherwise horizontal forces are not in equilibrium.

Edit: I don't believe moments are in equilibrium for Case 8 either. Looks like an error in the Air Force document.

Edit #2: Correction, see direction of forces and moments below.

BA

 

[Celt83]

steveh49:
oh yeah any real beam with those properties in that arrangement would violate Euler Bernoulli assumptions. My model was purely a numbers exercise.

 

[BAretired]

Kleinlogel Frame 31

Naming of nodes and members differs from Air Force document. Symbol J stands for moment of inertia (instead of I).

Coefficients

Case 5 and 6...corresponds to AF case 8 & 9

BA

 

[nargonaut7]

Greetings, everybody,

I apologize about my delay to reply, in the past two weeks I had to work mostly in another project, but managed to squeeze some time to do new calculations based on the replies I received here, for which I am grateful. But finally today I can post an answer, unfortunately not with good news.

I am attaching my calculations of a bracket example with easy dimensions and a load, all are completely different from mine, but they are easier for testing purposes. Initially, I thought that I was on the right track because, with the last clarification from BAretired, thanks for that, I was able to get the formulas I previously had of the Air Force Manual, Table 5-5, cases #8 and #9 to work well by doing a = 0 and d = 0, as s/he indicated, and you are right that Q is not defined there, I had also noticed that, but this doesn’t matter as you also wrote because d = 0 in this case. They work well because the combination of those two cases gave me a set of six reactions that made all Statics equations (summations of forces in X and Y, and of moments) of my situation with force on the overhang, equal to zero. To do that, I did a summation of moments for the inclined brace only, to obtain the unknown internal moment between this member and the horizontal one, and plugged this result with opposite sign in the summation of moments for the horizontal member, and the result was zero. Of course, that made me happy. Later, there was another post by BAretired, thanks also for this one, which mentioned a book by Kleinlogel, and I could download a PDF file of it from internet. It seem to be an excellent reference, the same as the AF Manual. The frames BAretired mentioned, 31/5 and 31/6, match my configuration, so there is no need to do the adaptation with a = 0 as with the Air Force Manual, but once again those two cases have to combined. I did the same as before with the six reactions to obtain the internal moment at the welded joint of the brace to the horizontal part, and again introduced this result in the summation of moments for this member, and got zero again. However, only the horizontal forces are the same between these two methods, the vertical forces and moments are different. I thought this was odd, as I always thought that the solution set of reactions for a Statics problem would be unique. But I thought that maybe there are two solutions for a statically indeterminate problem like this one. In the attachment I am showing the equations to verify the combined reactions from both methods. But, as I continued thinking about this, I did an experiment, I kept the horizontal forces, because the results are the same in these two methods, but put arbitrary vertical reactions and support moments that satisfy two rules: that the sum of the vertical reactions is 1,000 lb going up so they can oppose the external load, and also that the sum of the support moments is 2,000 lb-in clockwise so it can be added to the 22,000 lb-in from the external load to cancel the 24,000 (2000 x 12) lb-in CCW at the wall support of the brace. Initially, I tried with three combinations of reactions, all of them within the limits of the absolute values of the combined reactions obtained from the Air Force and Kleinlogel, and all of them worked, so I thought that those values could be the boundaries of the reactions, and this would be useful to calculate the stresses, but later I tried a set with all vertical forces and moments outside of those limits, and it worked also. So, I ended where I started, not being able to determine the reactions and moments. Both the Air Force and Kleinlogel work by themselves, but not getting matching results, and the infinite other solutions that also work, makes me call for help again. I am sure that I am making a mistake, but don’t know which.

Finally, I don’t want to send this post without acknowledging the reply by centondollar, thanks also for it. What you mentioned there about the eigenvalue of the matrices, I have to confess that I have seen that word “eigenvalue” before, but even with my Master in Mechanical Engineering I don’t know what that is. Maybe I should have researched this before posting this reply, but I didn’t want to delay it and then maybe on Monday I will have to work more on the other project. So, due to my ignorance, I cannot answer your comment, I can only say that I was simply using Cramer’s Rule, which is based on matrices, to solve a system of four equations with four unknowns I obtained by using the Slope-Deflection Method. To verify if this is possible, I used the Excel spreadsheet I prepared to solve these systems, and solved a 3 x 3 system of an example that is included in one of the two books I used to “learn”, I am still very shaky, this Slope-Deflection method. I am attaching here also that example and my Excel solution. So, I am guessing that I could use the same approach to solve my equations, the problem is that the determinant of the coefficient equation I wrote is zero, for sure that I am making a mistake but haven’t been able to identify which it is. But I am focusing more on the Air Force and Kleinlogel method at this moment, I think they would provide a more straightforward solution to include in the report, and again, as I mentioned in my first post, that report will include results from an FEA that will be run in a commercially available and reputable software. My other limitation is that, for my Master, a class of FEA was elective and I didn’t take it. I wonder if solving FEA by hand, or programming simple equations on Excel, equations that can be verified, a simple case like this one is possible, I couldn’t understand that from your answer. A mandatory class I took was on numerical methods, there I learned to solve partial differential equations (elliptic, parabolic, hyperbolic) with matrices that I programmed in Excel, but we didn’t touch FEA, and that was 18 years ago, and never used that again, so I don’t remember anything, other than solving for a taut vibrating string and for the distribution of temperatures on a plate.

Well, as I always say, sorry for writing so much, but I hope this is interesting, and I will appreciate very much any feedback.

And sorry that I am not answering to steveh49 and to Celt83, maybe the answer to my problem is there, but this is all the time I have today.

Thanks, and have a good day everybody!

 

[nargonaut7]

And here is the second file attachment

 

[BAretired]

@nargonaut7,

I am sorry you are having so much trouble with this problem, as it should be fairly straightforward. I think you deserve a medal for tenacity. I intend to have a closer look at the problem over the weekend (if someone doesn't beat me to it).

BA

 

[BAretired]

The results from moment distribution seem to agree with Kleinlogel. I have not checked the USAF formulas. Will do that tomorrow.

I am using joint symbols A, B, C, D in your first sketch, not as shown in subsequent sketches.

Using Moment Distribution
I/L for AC = 6/18 = 1/3; I/L for BC = 1.5/21.63 = 0.0693 ; Distribution Factor = 0.8278 and 0.1722 for AC and BC respectively
Mc = 4000"#; Mca = 3311; Mac = 1655.6
Mcb = 688.7"#; Mbc = 344.4
Ha = Hc = (4000 + 1655.6 + 344.4)/12 + 1000*18/12 = 500 from applied moment + 1500 from P acting at C = 2000#
Va = 3311*1.5/18 = 275.9#
Vb = 275.9 + 1000 = 1275.9#


BA

 

[centondollar]

Cramer´s rule is a method to solve a system of linear equations, and even though it includes taking determinants, it is not related to the eigenvalue problem (solving a homogeneous system of equations so that there are an infinite amount of solutions (continuous problem setting) or a countably finite amount of solutions (discrete problem setting)).

You may try Gauss elimination instead.

https://en.wikipedia.org/wiki/Gaussian_elimination

For a 3x3 or 4x4 system, I expect it to be more straightforward to use than Cramer´s rule.

 

[MIStructE_IRE]

TL;DR

So i’m going to pin the supports and the joint from the diagonal and allow the top horizontal member to cantilever over as shown.

Job done, keep it simple.

Moment in red, axial in blue.

 

[BAretired]

The 'K' term is incorrect in Table 5-5 in the USAF document (see below) so any values involving K will be wrong as well.

I agree, MIStructE_IRE that keeping it simple is best when designing a new bracket. But, when checking an existing bracket with rigid joints, this could be useful.

BA

 

[271828]

Let it go man. LOL. I have never in 30 years seen a simple problem overcomplicated to this extent. Not even a close second, and I have worked in R&D.

 

[BAretired]

Let it go man. LOL. I have never in 30 years seen a simple problem overcomplicated to this extent. Not even a close second, and I have worked in R&D.

I am happy to let it go, 271828. Pin joints were suggested from the start of the thread, but the OP needed a solution for a bracket with rigid joints. That solution was found using moment distribution and confirmed with Frame 31 of Kleinlogel. Air Force Table 5-5 did not confirm the result, suggesting that AF Table 5-5 should not be used.

If this thread annoys you so much, there is a very simple solution! Stop reading it!!

BA

 

[271828]

I was addressing my post to the OP.

You're right, though! Tuning out now.

 

[BAretired]

I can only say that I was simply using Cramer’s Rule, which is based on matrices, to solve a system of four equations with four unknowns I obtained by using the Slope-Deflection Method.

It is not necessary to solve four equations. Using Slope-Deflection equations, rotation at point C can be solved directly. Rotation at A and B are zero as they are fixed.

BA

 

[nargonaut7]

Hello everybody,

Once more I have to start my post apologizing about my delay, because in the last month and a half I had to work in other projects.

Well, thanks once more to BAretired, not only for your recent post verifying Kleinlogel using Moment Distribution Method (but later I indicate the questions I have about this), but also for your kind words about my tenacity. Actually, I think I am just doing my job, as I need to accompany the FEA report of this bracket with a manual calculation. So, I have an obligation to my employer, but you are helping me just because you want to, and the same goes to other contributors to this thread, so I am grateful to all.

And, I think I shouldn’t simply assume this bracket as pinned at its three supports (which is a calculation I had already done before my original post, and there I reported the maximum moment I got at the horizontal member) because I have never in my life seeing what are the results from the calculation of a closer approximation to reality, which is to assume those joints as fixed (I wrote “a closer approximation” because those three welded joints, if they don’t have lateral stiffening plates or gussets welded between members and between them and the steel wall, those joints will have certain flexibility). In other words, I cannot select between two approximations without comparing their results, at least for a similar case. And, I also think that complex calculations are not only for entry-level engineers, because calculations have to provide a degree of accuracy that is considered adequate depending on the particular situation, and if the required calculations to obtain the desired accuracy are complex, the assigned engineer, regardless of his or her level, has to either undertake them or look for help (like I am doing in this Forum, and anyway these Kleinlogel frame formulas are not complex, it is simply that I have never seen them before).

But well, back to the present calculations, after BAretired indicated that the K factor of the USAF Manual is wrong, I verified that, in the case of an applied moment, USAF Case (8) and Kleinlogel Case 31/5, with the K factor as defined by Kleinlogel, and using it at the USAF formulas (that is, with K = 0.208 instead of the original 0.375), the results match between both methods for reactions V and M, but they are still different in reaction H, and this is because the USAF manual has that factor K in both the numerator and the denominator of the formula, while Kleinlogel has that factor only in the denominator (“hidden” inside the N factor). So, I am not paying attention to the USAF Manual anymore, at least not for this bracket, even if it still seems an excellent reference for many other situations. I think that the discrepancies that BAretired has found in this Manual could be solved if we could check the reference that they mention, which is William Griffel’s “Handbook of Formulas for Stress and Strain”, but I haven’t found this book.

So, as I show in the attachment picture (I am sorry about its large size, we have to zoom to read it, and then it doesn’t fit in one screen), in Part 1, I was able to reproduce BAretired’s MDM calculations where I put red checkmarks, I did this in the Excel cells marked in green (the formulas I used are to the right). But I couldn’t follow cell B19, marked in orange, because the formula s/he used, with that 1.5 factor, is from Kleinlogel but for the V reaction, which is horizontal in our case, and the book formula is with the external moment when s/he used the internal moment. I am not saying that what s/he did is wrong, it is just that I don’t know where this formula comes from, because I thought that s/he was applying MDM but I have only found a similar formula in Kleinlogel. Actually, when I first read this post, I didn’t understand why the four moments have the values indicated there, but then I found the book indicated in Part 2 of the attachment, and there I studied the basics of the MDM method, I pasted the two key principles there (formulas 9.15 and 9.16), and with those now I understand those moment values. I even tried, but very clumsily, I recognize that, to apply the sequence of successive approximations of the MDM method to obtain moments Mac and Mbc as I also show in this Part 2.

Now I am more interested in the Moment Distribution Method, I found this method a few weeks ago in two books where I learned the basics of the Slope-Deflection Method, and there I learned (remember that I am mechanical, not civil or structural) that MDM was widely used until computers became available. It seems to me that both methods are general and powerful approaches compared to the specific, special-case formulas of Kleinlogel (and of the USAF manual). Several weeks ago I tried, unsuccessfully, to apply both methods to this bracket, but I didn’t put much effort at that time with the MDM, my problem was that the good examples for this method I could find were only for continuous beams with three or more supports (a situation that I was able to solve, 34 years ago, using the Three-Moment Theorem), so I couldn’t figure out how to apply MDM for this bracket.

So, after failing to understand all the results from BAretired, I tried, also unsuccessfully as I show in Part 3, to use the results from Kleinlogel to get four summations of moments (at the whole bracket, at the main horizontal member, at the brace, and around the joint main-member-to-brace) to be zero. Initially, I tried with the signs of the moments that I thought were correct, but that didn’t work, so I tried a couple other combinations, and then I decided to program the formulas for those four summations in Excel so I could try all 16 combinations of both signs of those four moments. But I never got all four summations to be zero at the same time. Actually, never got more than two to be zero at the same time, and the summation at the main member was never zero. In this approach, the moments at the joint are different, in contrast with what I did in Part 5, described later, where those moments are equal and of opposite signs, and Kleinlogel results for them (3311.2 and 688.8 lb-in) match those found with their corresponding moment distributions factors.

After I couldn’t make work the results from Kleinlogel, as I show in Part 3, I need to say that I am still confused by what I indicated in my previous post, which now I show again in Part 4 of the current attachment, with the same two examples of “arbitrary solutions” as before (but now they are less cramped than in the snapshot I attached on my previous post), together with the verification of the results from Kleinlogel. So, my problem is that there are, unless I am doing something wrong, many combinations of those four reactions (vertical and moment at two wall supports) that all satisfy the Statics summations of forces and moments. I mentioned in my previous post the conditions of the reactions to make these equations equal to zero, but now I numbered them in the attachment: “Condition 1” is that the horizontal reactions always have to be 2,000 lb. in opposite directions; the second condition is that the algebraic sum of the vertical reactions is 1,000 lb. going up (so they jointly cancel the external load); and the third one is that the algebraic sum of the support moments is 2,000 lb-in clockwise. So, to verify the Kleinlogel solution and for the two arbitrary solutions shown in this Part 4, I followed the same method for all three sets of reactions: after the six external reactions meet the three conditions described before (either calculated with Kleinlogel formulas or set arbitrarily), the summation of moments in either member only has the internal moment at the joint between them as a single unknown, solving for this internal moment and plugging it in the other summation with sign reversed makes the other summation equal to zero. But of course, this internal moment is the same and of opposite sign between these two members, so this is of course different to the distributed moment mentioned in Parts 1 and 3 (82.8 % of the external moment for the stiff horizontal member, 17.2 % for the flexible inclined brace). Besides those two arbitrary solutions shown here, I found several more, all matching the same pattern, but I always thought that the solution set of reactions for a Statics problem would be unique, so I am very confused about this.

And, Part 5, which is the last one of the attachment, shows my recent attempt to solve this bracket with the Slope-Deflection Equation. I tried this method several weeks ago, but now I tried it again because BAretired indicated that the “rotation at point C can be solved directly”. But I couldn’t do that and ended once again with four unknowns, and could only write three equations. Some weeks ago I wrote four equations but they were not independent so the determinant was zero, as I reported in a previous post. As I said earlier, I would like to learn this method because it seems more general than the Kleinlogel formulas.

I also want to use this posting to reply to three other contributors to this thread.

Thanks Celt83, for mentioning the “matrix analysis method”, I just did a quick search with these words and found that there are two approaches, the stiffness matrix and the flexibility matrix. So, that is more for me to learn.

Thanks centondollar, and yes, you are right that the Gaussian elimination needs less steps than Cramer to solve systems of equations. I used Gauss in one of my Master’s classes as I mentioned in a previous post, but that was 18 years ago so now I don’t remember how it goes, and to do repetitive calcs I would need to do a program, and I haven’t done any programing in several years, while Excel is very convenient because it calculates directly the determinants to use Cramer’s rule.

Thanks MIStructE_IRE for your post. In my original post, I discussed the result I got for the three-pinned bracket you suggested, and earlier today I mentioned this again. I know that this thread is long so it is easy to miss things here, so you can search in this page with CTRL F and “advice” (it will have less hits than other word searches), this will find the last paragraph of my original post, where I mentioned this result.

And finally, I also want to report that I found in Internet the other reference about frames that Roark has besides Kleinlogel, which is “Frames and Arches” by Valerian Leontovich. It is an excellent book of course, but it doesn’t have this triangular frame, he only has rectangular and trapezoidal frames, and arches.

 

[nargonaut7]

Sorry, I am not proficient yet attaching files. Here it is.

 

[rb1957]

how much ink is going to get spilled for this ?

if doing a hand calc, why FEA it ?

if you have to "validate your FEA", then why not show that the simple cantilever, or the simple pinned brkt works ?
but then repeat the previous question ... "surely" this simple brkt is a simple hand calc ??

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

 

[BAretired]

nargonaut7,
This is not a difficult concept.

Considering only the Moment Distribution Method, the 1000# load is shifted from D to C and a moment of 4,000"# is applied at point C. The effect of that combination is equivalent to the actual applied load at the end of the cantilever. The applied moment is resisted by two members, AC and BC in accordance with their stiffness. A carry-over factor places half the moment to the opposite end of each member. No need for any further distribution.

I agree that Hc was a typo. Should have been Hb.

From Moment distribution, Mac = Mca/2 (no further distribution is necessary as A is a fixed end).

Va = sum of moments on member AC/length = 3311*1.5/18 or (Mac+Mca)/18. (not from Kleinlogel)

Note that 3/2 in Kleinlogel is in agreement with 1.5 above. Note also that the vertical member CB in Kleinlogel corresponds to the horizontal member AC in your sketch. In other words, Kleinlogel must be rotated 90 degrees to compare coefficients.



BA