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Author | Topic: A test of your common sense | |||||||||||||||||||||||||||||||||
Taz Member (Idle past 3321 days) Posts: 5069 From: Zerus Joined: |
You know, you remind me of the following scene from doctor strangelove.
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xongsmith Member Posts: 2587 From: massachusetts US Joined: Member Rating: 6.4 |
PD writes:
Assuming whatever the beam is sitting on is stable and won't move, I would say it will fail at B and C. Yes, those are good assumptions, given the dearth of info Taz gave us. But I suspect he wanted us to think they would move. Why draw them the way he did if not to emphasize the differences in each supporting side? A little reverse detective work........- xongsmith, 5.7d
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xongsmith Member Posts: 2587 From: massachusetts US Joined: Member Rating: 6.4 |
Lol thanks for that!
- xongsmith, 5.7d
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Taz Member (Idle past 3321 days) Posts: 5069 From: Zerus Joined: |
I hope to god you're not an engineer. Here's why. Yes, the world should be based on good solid engineering foundations. But if you are as dense as that soldier in that video clip I posted in my previous post, then you are a useless engineer. I've run into guys like that at work. Sure, they graduated with very high gpa's. But you will very quickly discover while working with them that they can't really apply what they learned to real life. You have to be a lot more flexible than that. My current line of work doesn't have the luxury of having all the known factors neatly spelled out like what you find in text books. Just look at the drawing. Don't over think it.
Nuff said about that.
quote:Actually, there are 2 reasons why I put a roller in there. (1) It's out of habit. (2) I didn't want people to be concerned with the x axis.
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xongsmith Member Posts: 2587 From: massachusetts US Joined: Member Rating: 6.4
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If it were even remotely possible to assign myself with anyone in that funny snippet (thanks again for that!!), it would be the Peter Sellers character. But - actually the snippet was not really relevant to this thread. Sorry.
Try replying to Message 14 first, if you are set on blowing off Message 3 and Message 4 by NWR. You say:
Actually, there are 2 reasons why I put a roller in there. (1) It's out of habit. (2) I didn't want people to be concerned with the x axis. Well, see? there: you are now calling it a "roller"! That is information that I had to assume with probabilities less than desirable - you could have just simply said that at the outset. Habit? Weird you. Thumbs up! And if you think the x-axis is nothing to be concerned about, maybe you should restart your algebraic physics problem knowledge at this time. Is the roller going to only move in the Y-axis??? Read Message 14. See? I think you thought you had a cool simple little thing here, only it was way more complicated than you first thought. Clue: if I had all the info, it would still be WAY TOO HARD FOR ME! You say:
Just look at the drawing. Don't over think it. The drawing is very bad. No way to sugar-coat it for you. It's bad. Edited by xongsmith, : Forgot Edited by xongsmith, : No reason given.- xongsmith, 5.7d
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frako Member (Idle past 335 days) Posts: 2932 From: slovenija Joined: |
The beam would fail at point D. or should i say loose support at point D the circle supporting it would shoot out when the beam would bend do to the 2 forces acting on it. Assuming the triangle support is inside a niche in the beam as shown on the diagram. If it is not and it is just resting on top of the triangle support then both points A and B would fail as the beam bends do to the forces applied to it. The triangle would topple and the circle would shoot out. Or part A would slide down the triangle and beam would slide moving the circle inwards but if force is still apllied after that happened the circle would still shoot out.
Edited by frako, : No reason given.
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Tangle Member Posts: 9514 From: UK Joined: Member Rating: 4.8 |
Not enough information.
If the beam is flexible it would simply bend between the two endpoints and eventually squeeze past them. If the two end points aren't fixed, they'd be forced out sideways as some of the downward force became lateral by the bending of the beam. If the beam is not perfectly rigid and the two endoints are uncrushable and imoveable, the beam would break at either or both points b & c as the beam flexes down or mid centre between a & b depending on the amount of bending the beam allows itself (ie if if deforms into an arc) Assuming a rigid beam and fixed and uncrushable end points, the beam could fail at any point though an unevenness or weakness in it's structure. Assuming a perfectly uniform and rigid beam with fixed and uncrushable end points points (was this your meaning?) then I'm not sure. I would guess near to or at A as both downward pressures would be exerted evenly between a & b but there would be a penetrating force at A combined with pressure at a point.Life, don't talk to me about life - Marvin the Paranoid Android
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dwise1 Member Posts: 5952 Joined: Member Rating: 5.2 |
Assuming a rigid beam and fixed and uncrushable end points, the beam could fail at any point though an unevenness or weakness in it's structure.
I think that's the point of this topic. You might as well have started with the classic "Assume a spherical cow ..." For example, back in the late 1970's, a problem was presented and a variety of four outcomes were presented to the poll respondents. The problem as that a ball was descending down a spiral track. The variety of outcomes all had to do with what would happen to the ball's trajectory once it left that spiral track. The correct choice, that the ball would proceed straight out and fall into a trajectory consistent with free-fall was not the winner. Rather, the choice that the ball would continue to follow a spiral trajectory as it fell was the winner. And completely wrong. The question is what the "common sense" solution would be. Even though that would invariably be wrong.
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RAZD Member (Idle past 1434 days) Posts: 20714 From: the other end of the sidewalk Joined:
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Hi Taz,
Taking this as a standard engineering load diagram for "pinned end" loading and assuming no distributed load for the weight of the beam:
The deflection will be greatest in the middle, following the next order of curves (bending is the integration of the sheer curve, deflection is the integration of the bending curve). Combined sheer and bending stress will be greatest at the load points, B and C. With a distributed load for the beam weight, there would be added sheer and bending loads:
Thus the greatest sheer load is at the ends, and the greatest bending stress is in the middle. Where the beam fails depends on the relative amounts of P and w and whether it is more sensitive to failure in sheer or bending, and of course the condition of the beam and the end foundations .... and whether or not a butterfly farts in Mexico .... This is based on what I know from my education and experience, as much as I know 1+1=2, and this is why my "common sense" is likely different from yours ... and why there really is no "common" sense of things at all. A better term would be common ignorance ... Enjoy. Edited by RAZD, : added Edited by RAZD, : /list Edited by RAZD, : No reason given. Edited by RAZD, : ...by our ability to understand Rebel American Zen Deist ... to learn ... to think ... to live ... to laugh ... to share. Join the effort to solve medical problems, AIDS/HIV, Cancer and more with Team EvC! (click)
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Percy Member Posts: 22504 From: New Hampshire Joined: Member Rating: 4.9
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Hi Taz,
I'm curious about the answer, but I'm not sure this has much use as an illustration of common sense gone awry. A good illustration would be one where most people have the same common sense reaction. The problem with your example is that it probably draws a random mix of reactions from "the middle" to "points B or C" to "the ends" to "beats me" (my personal favorite). I like the Monty Hall example of common sense gone awry, but the proof is so abstruse that few understand or accept it. I wonder if there's a simple example where the correct answer is easily understood. --Percy Edited by Percy, : Minor correction.
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Taz Member (Idle past 3321 days) Posts: 5069 From: Zerus Joined: |
Good god. Don't over think it, people. Don't say things like "depends on if the beam is perfectly even blah blah blah". And also don't try to insert math or engineering principles. The point of this thread is common sense. Just by looking at it, where does your common sense tell you the beam/plank/piece of wood/person/car/whatever will fail? We'll see if your common sense agrees with what reality actually is.
As I tried to say before, common people often have a mistrust in scientists and engineers (and other professionals in their respective fields) because some things don't fall in line with people's common sense. Take that simple physics experiment, for example, where college students were asked to roll a ball through a curved path. Most people tried to roll the ball in a curved path. You know what I'm talking about. That's an example of where common sense would fail most people. Again, don't over think it. Just by looking at the diagram, where does your common sense tell you the long thing will break given enough load?
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Tangle Member Posts: 9514 From: UK Joined: Member Rating: 4.8 |
Again, don't over think it. Just by looking at the diagram, where does your common sense tell you the long thing will break given enough load? Um, what has common sense got to do with it? It's an engineering question using engineering language. You're asking people that think, not to think. In my opinion, thinking IS using common sense. Relying only on common sense is a last resort equivalent to saying "I don't know but my E. coli tell me......." If I ever need to remind myself of that, I go read something rediculous about quantum theory - or the law. Edited by Tangle, : No reason given.Life, don't talk to me about life - Marvin the Paranoid Android
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Coyote Member (Idle past 2135 days) Posts: 6117 Joined: |
Good god. Don't over think it, people.
OK. The bloody thing will break in the middle. Now get on with it.
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frako Member (Idle past 335 days) Posts: 2932 From: slovenija Joined: |
Again, don't over think it. Just by looking at the diagram, where does your common sense tell you the long thing will break given enough load? If the supports dont matter then the beam will brake at the 2 middle points i think b and c if the force applied to those points is equal. The beam will look something like this afterwwards \___/
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ringo Member (Idle past 441 days) Posts: 20940 From: frozen wasteland Joined: |
What we have here... is failure to communicate.
And you're going to have to answer to the Coca-Cola company for that. My common sense - and if I had any, I probably wouldn't be wasting my time on this topic - tells me that the beam will break at B and C at the same time. So let it be written. So let it be done.
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