A funny statics problem 13

[NedGan76]

Hi, everyone!

Here is an interesting and funny statics problem, to exercise engineering brains and cheer up.

Given the following structure can you tell, without calculating, in which direction point A will move after deformation:
A) up
B) down
C) will not move

... and why? Point A is not a hinge. No structural analysis programs allowed.

 

[Tomfh]

I think it's the same deflected shape? The way I'm seeing it, the deflected shape is the horseshoe shape. The roller simply rotates that shape clockwise by a small amount, which raises B in the process. Like this:

 

[BridgeSmith]

I just realized it's trick question. The title states this is a statics problem. In a statics problem, nothing moves, so the correct answer is C. The original shape is the shape with the loads already applied.

 

[Just Some Nerd]

Yea, that is the MF for me. If there is no reaction in the left column, how is it affecting the deflected shape?

In ""reality"", what's happening is that there is a temporary reaction that rotates the system until there is no longer a reaction. That's my interpretation at least and makes the most intuitive sense.

----------------------------------------------------------------------

Why yes, I do in fact have no idea what I'm talking about

 

[NedGan76]

BAretired, this is a good approach to compare the three cases.

Tomfh, this is a good explanation. Without the roller support at bottom left, the system is in an unstable balance. An infinitely small force is needed to rotate the structure about the fixed support.

BridgeSmith: "In a statics problem, nothing moves".
Please, allow me to kindly disagree with this. In statics, nothing moves as a free body - otherwise it is kinematics. But different points of the structure move due to deflection of members. It is also important that they move very slowly. Otherwise, it is dynamics.

Cheers!
Ned Ganchovski

A better software for your calculation notes, free and open source:

GitHub - Proektsoftbg/Calcpad: Free and open source software for mathematical and engineering calculations.

Free and open source software for mathematical and engineering calculations. - Proektsoftbg/Calcpad

github.com

 

[XR250]

I think it's the same deflected shape? The way I'm seeing it, the deflected shape is the horseshoe shape. The roller simply rotates that shape clockwise by a small amount, which raises B in the process. Like this:

Now it makes sense to me - thanks

 

[phamENG]

Yeah - the 'temporary' reaction and time history idea make the most sense. It's important to remember that all of our deflected shape sketches are grossly over-exaggerated. If they deflected that much, then the deflections would be enough that the forces would no longer be truly equidistant from the center support and there would be a secondary net moment resulting in a real reaction at the roller.

 

[Eng16080]

NedGan76,
I think the problem should state that:
[ol 1]
[li]The members are weightless, and[/li]
[li]As the horizontal member deflects, the horizontal length (in the x direction) is assumed to stay the same.[/li]
[/ol]

If those two conditions aren't met, then I think it's impossible for the horizontal reaction bottom left to be zero.

Assuming the two conditions above are true, I picture a propeller with a couple of weightless legs attached. The propeller can be turned to any angle and it's still stable. So, if it's turned until the left leg happens to be where the support is (directly below point A), that's the solution.

Thanks for such a cool problem to think about. This was really working my limited brain cells.

 

[Denial]

Eng16080.[ ] Given the context of NedGan76's description of the problem, I think that weightless members and small displacement theory could be taken for granted.[ ] No harm in spelling them out, but no need either.

NedGan76.[ ] How did you encounter / conceive this interesting little problem?[ ] It certainly attracted (and tested) some of the site's heavy hitters.

[sub][ ]—————————————————————————————————[/sub]
[sup]Engineering mathematician / analyst.[ ] See my profile for more details.[/sup]

 

[JLNJ]

Edited as I try to apply the teeter-totter theory to the system.

 

[Retrograde]

Please, allow me to kindly disagree with this. In statics, nothing moves as a free body - otherwise it is kinematics.

Allow me to kindly disagree with this. The member has moved as a free body - it has rotated about the top central support.

I stand by my previous post - it is unstable when considering small displacement theory.

 

[NedGan76]

Retrograde. I would even more kindly disagree with this either. There are three types of structures:
1. kinematic;
2. statically determinate;
3. statically indeterminate.

This structure above is definitely "2. statically determinate" and it is stable. A body has three DOF's in the plane - 2 translations and 1 rotation. So, you need three restraints that are not colinear to get it fixed. This is satisfied in our case. If you see my previous post, I already solved it by hand and found the respective support reactions. So, it moves due to the deflection of members. It would become unstable, if we take out the bottom sliding support. Then, it will be able to rotate around the top support as a free body. Now, it cannot.

Denial. I know this from more than 25 years from my professor in the university. He was a great mind and remarkable person. I have some more like this problem in the pocket. I may post another one in the future.

Cheers!
Ned Ganchovski

A better software for your calculation notes, free and open source:

GitHub - Proektsoftbg/Calcpad: Free and open source software for mathematical and engineering calculations.

Free and open source software for mathematical and engineering calculations. - Proektsoftbg/Calcpad

github.com

 

[GeorgeTheCivilEngineer]

"So, you need three restraints that are not colinear to get it fixed. This is satisfied in our case"

Is it? There's only a single vertical support and a single horizontal support.


Edit - sorry top one is pinned.

Ignore me. Lots of replies to catch up on.

 

[NedGan76]

GeorgeTheCivilEngineer. No, there are two horizontal supports and one vertical. The top middle support provides restraint in two directions: X and Y. Yes, it is pinned. Sorry, lost in translation. Actually, this is very basic stuff. I am a little surprised that we have to discuss that here.

Cheers!
Ned Ganchovski

A better software for your calculation notes, free and open source:

GitHub - Proektsoftbg/Calcpad: Free and open source software for mathematical and engineering calculations.

Free and open source software for mathematical and engineering calculations. - Proektsoftbg/Calcpad

github.com

 

[human909]

Thanks for the little problem that has stumped many.

Actually, this is very basic stuff. I am a little surprised that we have to discuss that here.

It is fundamental to structural engineering analysis, but that doesn't mean it is "basic", simple or even necessary.

A good grasp of the theoretical underpinnings of static determinacy is not really that important in everyday applied structural engineering or even advanced applied structural engineering. There are no doubt some very good engineers who don't grasp these concepts well and also some poor engineers who do.

 

[GeorgeTheCivilEngineer]

GeorgeTheCivilEngineer. No, there are two horizontal supports and one vertical. The top middle support provides restraint in two directions: X and Y. Yes, it is pinned. Sorry, lost in translation. Actually, this is very basic stuff. I am a little surprised that we have to discuss that here.

I'm a bit perplexed as to why you replied - I edited my post 3 minutes after I posted and struck out that bit...

 

[WesternJeb]

Ned, this is a fantastic college-level question to get young minds thinking, similar to how human909 put it. Very challenging and fun, but not much practical application that we are used to seeing.

 

[Denial]

It also cranked my OLD brain up.

[sub][ ]—————————————————————————————————[/sub]
[sup]Engineering mathematician / analyst.[ ] See my profile for more details.[/sup]