3D reaction force calculation 2

[mdc1973]

Hello,
I wish to calculate the reaction force at Cz to keep the structure in equilibrium.

The linked diagram shows a platform that weighs 500kg.
The platform has a point load 1.9KN.
The platform size is 2.8m long x 0.9m wide and it is restrained by two shafts on opposing corners, free to rotate axially.

Link

 

[BAretired]

Correction:

Joint B has 3 constraints, Fx, Fz and φy.
should read:
Joint B has 3 constraints, Fy, Fz and φy.

BA

 

[ThomasH]

BAretired:

Then perhaps you can understand my question. The figure has five stated constraints. You mentioned five constraints, but not those five. Instead you have "added" two based on your interpretation of the figure. And also skipped two that are in the figure . Hence, the question.

In your first post I believe you had the same interpretation as I have. Then you changed it for some reason. Maybe your latest interpretation is correct, maybe not. Only the OP can clarify that.

But I disagree on your statement that it can't be analyzed by statics alone. It is a static problem.

Friendly Regards

Thomas

 

[dhengr]

This has been an interesting thread, with some of E-Tip’s sharpest members participating, with some slightly different interpretations of what’s really going on. What distresses me lately is that a fairly large percentage of the OP’ers. who come here don’t understand their own problem well enough to explain it in a meaningful way to other smart engineers. They leave us guessing. They’ve got some wild assed Rube Goldberg idea with no idea how it works, and actually think that is good engineering. And yet, they claim/pretend to be engineers or capable technical people. The number of poorly defined problems and questions that are presented here is really scary. Too many of the members seem to be designing and building a bunch of crap for the general public without even being able to define what they are doing, how it really works or why that strange arrangement is needed, and still they are pushing that stuff out the door, and thinking that’s good, and safe design.

 

[BAretired]

ThomasH,

In my first post, I recognized that, due to the two hinges about the X axis, it was possible to solve for the reaction at A which I found to be P/3. I could see that the sum of reactions at B and C had to be 2P/3 and because P was halfway between the two, I erroneously jumped to the conclusion the BC reaction was split evenly between B and C.

I then realized that, not only was I wrong but it was not possible to calculate the split between B and C because there were unknown moments acting on A and B which could not be solved by statics alone. IDS came up with a solution by considering some assumed properties of the plate. As you can see, his analysis for the point load is substantially different than his later analysis using your assumptions.

And finally, it may be necessary to ask the OP what he meant to say but it is not necessary to ask him what he did say. His first post left no room for guesswork.

BA

 

[ThomasH]

BAretired:

Regardless of if my interpretation or your interpretation of the constraints is correct, the problem seems to me to be static. It may be indeterminate but that does not make it less static. I can't see any need to use dynamics.

As for the correct boundary conditions, we may never know since the OP seems to have left the discussion .

Regards

Thomas

 

[BAretired]

ThomasH,

Nobody is suggesting that dynamics must be used. The problem is statically indeterminate which means compatible deformations must be considered in order to arrive at a solution.

A beam with fixed ends is also statically indeterminate by definition.


BA

 

[rb1957]

if there is no My restraint at A and B then it is statically determinate.

if there is (My restraint at A and/or B) then it is statically indeterminate (just as a doubly cantilevered beam is statically indeterminate, or "hyper-static" in today's lingo). it does seem odd (but not wrong) that these fixed end moments are penalising redundancies, as shown by the increase in Cz based on the posted FEA results.

Quando Omni Flunkus Moritati

 

[ThomasH]

BAretired,

I know exactly what it means.

It was not me who stated "could not be solved by statics alone". That was the reason for my comment.

I'll think I will just leave it for now. We'll see if the OP comes back and if he has any comments.

Regards

Thomas

 

[BAretired]

rb1957,

if there is no My restraint at A and B then it is statically determinate.

if there is (My restraint at A and/or B) then it is statically indeterminate (just as a doubly cantilevered beam is statically indeterminate, or "hyper-static" in today's lingo). it does seem odd (but not wrong) that these fixed end moments are penalising redundancies, as shown by the increase in Cz based on the posted FEA results.

I agree with you except for your example (a doubly cantilevered beam is statically determinate), but I get the idea. As for the penalising of Cz, it made only a slight difference, presumably because My at A was greater than My at B. However, it made a huge difference to the magnitude and sign of the moments within the plate itself.

ThomasH,

Are we still having a friendly discussion? Perhaps I should have said "could not be solved by the equations of statics alone". Those equations state that the sum of forces and moments in each principal direction must add up to zero.

BA

 

[rb1957]

is that the sound of hairs being split ?

smile, remember the adage about arguing with an engineer and mud wrestling with a pig ...
or was it arguing with a pig and ...

Quando Omni Flunkus Moritati

 

[ThomasH]

BAretired,

We are still friendly (I hope). But I think you perhaps should be open to the possibility for other interpretations than your own. I thing the tread proves that there is more than one possible interpretation for the boundary conditions.

And:

You have six equations for static equilibrium. They may or may not be enough depending on the boundary conditions. If they are not enough you add criteria for deformations, based on (again) static assumptions. They are all static equations, some for equilibrium others for deformations. Correct?

You can add dynamics but in this case if would be very redundant.

And I can agree with rb1957 that it is a bit of splitting hairs. But since the way of communication in this case is written posts I can only comment on your written comments, do I agree or not. And when you say that statics isn't enough, I disagree. But I hope we still can be friends.

rb1957:
I think the line is: Arguing with an engineer is a bit like mud wrestling with a pig. After a while you realize that he likes it .

Friendly Regards

Thomas

 

[rb1957]

agreed (with the aruging/wrestling line) ... i was trying to have a little fun with it ...

failed again, sigh

Quando Omni Flunkus Moritati

 

[ThomasH]

rb1957:

I apologize for ruining your joke, I missed that one.

But I agree with the statement. 50+ posts for a fairly simple problem, that is a bit excessive.

Regards

Thomas

 

[IDS]

But I agree with the statement. 50+ posts for a fairly simple problem, that is a bit excessive.

As often here, the back-story is more interesting than the original question.

I do think it is interesting (and instructive) that some think the ambiguous diagram in the OP is quite unambiguous, especially since there are (at least) two different versions of what it unambiguously shows. FWIW, I think the most likely interpretation is that rotation about the Y axis is intended to be restrained, since rotation about the X axis is explicitly stated to be released. In any case, in any real structure the % restraint would be something between 0 and 100, so we should look at both cases.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[paddingtongreen]

ThomasH, if you really think that repeating that it is a statically determinate problem will make it so, no matter what, you are mistaken.

If those are trunnions at A and B, that is, restrained in all directions and restrained from rotation about the vertical and Y axes, the result is in the platform, unless it is rigid within itself, but we are structural engineers so we know that cannot be.

Dynamics are not involved here, they are not required to make a structure statically indeterminate.

OPs frequently try to simplify a problem and in doing so change it. When the answers do not suit, more info emerges. The fact that the OP in this case had to ask seems to imply that there is more than the simple problem we all answered at the beginning.

Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin

 

[ThomasH]

BAretired, IDS, paddingtongreen,

Ok, I will try to explain my reasoning and perhaps also apologize.

First, I completely that repeating a wrong answer does not make it correct. That is obvious.

Now if we look at the structure. It is a rectangular plate supported in three corners. Regardless of the support conditions it is a fairly simple problem. I would give it to any engineer on his/her "first day" at work and expect it to be solved. Either "by hand" or by using a computer.

It does not require any dynamics, on that we agree. And perhaps I was a but to stubborn regarding the statics so I just apologize.

As for the support conditions. Perhaps you are correct, I don't know. But I have seen this type of supports in a figure before (in an old textbook). And this leads to my final observation. This is the OP first post, on his first day of membership. This was one of my first observations since I thought the question was fairly simply. So there is no history. Student posting, perhaps. Again, I have no idea.

Now if I have irritated anybody, I apologize. That was never my intention.

Best Regards

Thomas

 

[BAretired]

ThomasH,

There is no need to apologize, so far as I am concerned but if you are apologizing, then apology accepted. It is probably time to put this issue to bed.

Best Regards,

BA