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.

 

[BAretired]

Slope Deflection Equations for moments at C

1. Mca = Mfca + (2EI/L)(2θc+θa)
2. Mcb = Mfcb + (2EI/L)(2θc+θb)

Since Mfca = Mfcb = 0 (no fixed end moments in these spans)
and since θa = θb = 0 (fixed ends);

3. Mca = 4EI*θc/L = 4E*6*θc/18 = 1.333E*θc
4. Mcb = 4EI*θc/L = 4E*1.5*θc/21.63 = 0.277E*θc

5. Mcd = Mca + Mcb = 1.61E*θc = 4000"#
6. Mca = 1.333 * 4000/1.61 = 3312"#
7. Mcb = 0.277*4000/1.61 = 688"#

Results are in agreement with Moment Distribution Method and Kleinlogel table.

BA

 

[nargonaut7]

Greetings, everybody,

Once again I have to start with an apology for another month and a half of silence, I have been busy with other projects and tasks.

Thanks once more to BAretired for the verification that three methods: Moment Distribution, Slope Deflection, and Kleinlogel book formulas, all agree in the same reactions and moments, which I represented in the attachment of today, even that it still has one question. I have learned in these posts and studying books about these three approaches, all of them were new to me (for those readers that are new to this thread, I am not Civil or Structural but a Mechanical Engineer). In summary, I understand that MDM gives us all the moments, internal and external, while Kleinlogel (by combining two cases in this situation) gives us the external ones and Slope-Deflection the internal ones, then we can calculate the others with Statics.

Initially I couldn’t understand why BAretired, in the post of March 21, when applying the Slope-Deflection Equation, considered the deflection at C (the joint between the two members) equal to zero, but after a lot of thinking I realized that the deflection of the inclined member is mostly due to axial deformation, and this effect is neglected in this method, this fact simplifies greatly the equations (in my attempt with this method back in January and February I got a system of 4 equations that were not independent, the determinant was zero).

But, even with the positive summary I have so far today, I still have a problem, which is that one of the three summations of moments is not zero, as I show in the attachment. I am sure I am making a silly mistake, but I have reviewed it several times and can’t find it. As always, I appreciate the help.

And, to address the impatience of rb1957, of course that I wanted to have this done a long time ago. Your two questions of March 21 were answered in my initial post, because I knew from the start that those two questions were going to be asked. But let me say it here in another way: I think I can go with a simplification (in this case, all three joints pinned) only after I have solved, at least once so I can compare results, the other end of the spectrum (which is all joints fixed; the reality in my case will be an intermediate rigidity, because we won’t have stiffening plates in those three welded joints, only continuous fillets all around the two square tubing pieces at all three joints). This is my plan and I will stick to it, others may go directly with the simplification because they have already done that comparison in the past, or because they don’t know how to solve the all-fixed case, but I don’t judge this second group. I only want to learn.

Thanks once again for the help and the time reading these posts.

 

[BAretired]

You are including the 4000"# moment twice. The applied moment is 4000"# which causes tension in the top of member AC, a clockwise moment. It is shown correctly in the diagram to the left.

BA

 

[nargonaut7]

Thanks BAretired, for your continued support.

But sorry, I don’t understand your reply, because the summation of moments that I called “AC”, is for the entire horizontal member. Actually, I should have labeled the point of application of force F as “D”, to call this summation “AD”. So, there are three vertical forces applied at this member, but only two of them create moment if the base point of the summation is either A or C, plus the two other moments (one internal, one external) so, I could be wrong, but I think that this summation should have four terms. Even if I break this member at C, I think I still have to apply that moment of  4000 lb-in at that point.

And, regarding the direction of the moments in the small diagram of joint C, I thought that the moments of  3311.2 and  688.8 lb-in, which “appear” on the left side of joint C based on their corresponding MD factors from the unbalanced moment of 4000 lb-in, that those two negative moments are applied by the joint to the horizontal and inclined members, but then these two members would apply an equal moment of opposite direction (positive, in this case) back to the joint. Then, those three moments at this diagram I drew cancel each other.

I added another diagram to what I posted yesterday with what I wrote above, to have everything in one place.

Thanks once again. I am optimistic that we are reaching the conclusion of this thread. The horse is smelling the barn!!

 

[BAretired]

Consider member ACD reactions. Mac is a reaction and must be used to check equilibrium of the member. Mca does not affect equilibrium of the beam; it is internal and is not used to establish equilibrium. Mcb is a moment applied to member ACD and affects equilibrium. Mcb is clockwise when applied to member BC, but is counterclockwise when applied to member ACD (reaction is always opposite to an action).

I hope this thread has helped in your understanding of the statics of a bracket with fixed joints.

BA

 

[BAretired]

I don’t understand your reply, because the summation of moments that I called “AC”, is for the entire horizontal member. Actually, I should have labeled the point of application of force F as “D”, to call this summation “AD”. So, there are three vertical forces applied at this member, but only two of them create moment if the base point of the summation is either A or C, plus the two other moments (one internal, one external) so, I could be wrong, but I think that this summation should have four terms. Even if I break this member at C, I think I still have to apply that moment of  4000 lb-in at that point.

There is an important difference between isolating members 'AC' and 'ACD'. In the latter case, the 1000# load acts on the isolated member at point D; in the former case, some of the moment from that load acts on member 'BC'.

When performing an equilibrium check on a member, all loads and reactions must be considered. The sum of vertical and horizontal forces and the sum of moments on the entire member must be zero, not just point 'C'.

BA

 

[ThomasH]

Hello nargonaut7

I have followed this thread on and of for some time now. I have not read all posts in detail but there is one question that i have that bugs me. What is the purpose of this "exercise"?

Thomas